Class 11 Maths Ncert Solutions Chapter 10 Ex 10.3 Straight Lines PDF

Class 11 Maths Ncert Solutions Ex 10.3

Class 11 Maths Ncert Solutions Chapter 10 Ex 10.3

Q-1. Write all the possible equations for the two axis, which is x- axis and y- axis.

Solution.

As we know that,

The x- coordinates at all the points on the y- axis is 0.

So, the equations on the y – axis is y = 0.

The y- coordinates at all the points on the x- axis is 0.

So, the equations on the x – axis is x = 0.

Q-2. Write the equation for the line which passes through the point (-5, 4) with having slope 14$\frac{ 1 }{ 4 }$.

Solution.

It is known that the equation for the line passing through any point, say, (x0, y0), with having slope m, is given by-

(y – y0) = m (x – x0)

Therefore, the equation for the line passing through (-5, 4) with having slope 14$\frac{ 1 }{ 4 }$, is

(y – 4) = 14$\frac{ 1 }{ 4 }$( x – ( -5 ))

(y – 4) = 14$\frac{ 1 }{ 4 }$( x + 5 )

4(y – 4) = (x + 5)

4y – 16 = x + 5

i.e., x – 4y + 11 = 0

Hence, the equation of the line which passes through (-5, 4) is given by-

x – 4y + 11 = 0

Q-3. Write the equation for the line which passes through the point (0, 0) with having slope S.

Solution.

It is known that the equation for the line passing through any point, say, (x0, y0), with having slope m, is –

(y – y0) = m (x – x0)

Therefore, the equation for the line passing through (0, 0) with having slope S, is

(y – 0) = S (x – 0)

i.e., y = Sx

Hence, the equation of the line which passes through (0, 0) is given by-

y = Sx

Q-4. Write the equation for the line which passes through ( 3, 33$\sqrt{ 3}$ ) which is inclined on x- axis at an angle 75$75^{\circ}$.

Solution.

The slope of the line inclined on the x- axis at an angle 75$75^{\circ}$ is m = tan 75$75^{\circ}$

m=tan(45+30)$\Rightarrow m = tan\left ( 45^{\circ} + 30^{\circ} \right )$ m=tan45+tan301tan45.tan30$\Rightarrow m = \frac{ tan 45^{\circ} + tan 30^{\circ} }{ 1 – tan 45^{\circ} . tan 30^{\circ} }$ m=1+1311.13=3+13313$\Rightarrow m = \frac{ 1 + \frac{ 1 }{ \sqrt{ 3 } } }{ 1 – 1.\frac{ 1 }{ \sqrt{ 3 } }} = \frac{\frac{ \sqrt{ 3 } + 1 }{ \sqrt{ 3 } }}{ \frac{ \sqrt{ 3 } – 1 }{ \sqrt{ 3 } } }$ m=3+131$\Rightarrow m = \frac{ \sqrt{ 3 } + 1 }{ \sqrt{ 3 } – 1 }$

It is known that the equation for the line passing through any point, say, ( x0, y0 ), with having slope m, is –

(y – y0) = m (x – x0)

Therefore, the equation for the line passing through ( 3, 33$\sqrt{ 3 }$ ) inclined at an angle of 75$75^{\circ}$ on the x- axis is given by:

( y – 33$\sqrt{ 3 }$ ) = 3+131$\frac{ \sqrt{ 3 } + 1 }{ \sqrt{ 3 } – 1 }$ ( x – 3 )

( y – 33$\sqrt{ 3 }$ ) ( 3$\sqrt{ 3 }$ – 1 ) = (3$\sqrt{ 3 }$ + 1 )( x – 3 )

y (3$\sqrt{ 3 }$ – 1 ) – 33$\sqrt{ 3 }$( 3$\sqrt{ 3 }$ – 1 ) = x ( 3$\sqrt{ 3 }$ + 1 ) – 3( 3$\sqrt{ 3 }$ + 1 )

( 3$\sqrt{ 3 }$ + 1 )x – ( 3$\sqrt{ 3 }$ – 1 )y = 33$\sqrt{ 3 }$ + 3 – 9 + 33$\sqrt{ 3 }$

( 3$\sqrt{ 3 }$ + 1 )x – ( 3$\sqrt{ 3 }$ – 1 )y = 63$\sqrt{ 3 }$ – 6

i.e., ( 3$\sqrt{ 3 }$ + 1 )x – ( 3$\sqrt{ 3 }$ – 1 )y = 6( 3$\sqrt{ 3 }$ – 1 )

Hence, the equation of the line which passes through ( 3, 33$\sqrt{ 3 }$ ) is given by-

( 3$\sqrt{ 3 }$ + 1 )x – ( 3$\sqrt{ 3 }$ – 1 )y = 6( 3$\sqrt{ 3 }$ – 1 )

Q-5. Write the equation for the line which intersects x – axis at a distance 4 units away from the left side of the origin having slope -3.

Solution.

We know that,

If a line having slope m intersects x- axis at a distance d, then the equation of such lines is given by:

y = m(x – d)

Thus, The equation of the line intersecting x- axis at a distance 4 units from the left side of the origin, i.e., d = -4, and having slope m = -3, is given by:

y = -3[x – (- 4)]

y = -3x – 12

Therefore, the equation of the line required is given by:

3x + y + 12 = 0

Q-6. Write the equation for the line which intersects the y – axis at a distance 3 units above the origin which makes an angle 60$60^{\circ}$ along the positive direction of the corresponding axis, i.e., x- axis.

Solution.

We know that,

If a line having slope m makes a y- intercept c, then the desired equation for the line is given by:

y = mx + c

Since, c = 3 units and makes an angle of 60$60^{\circ}$ along the positive direction of the x- axis, i.e.,

m = tan60$60^{\circ}$ = 3$\sqrt{ 3 }$

Therefore, the equation of the line required is:

y = 3$\sqrt{ 3 }$x + 3

i.e., 3$\sqrt{ 3 }$x – y + 3 = 0

Q-7. Write the equation of the line passing through the two points (-2, 2) and (3, -5).

Solution.

We know that,

m=y2y1x2x1$m = \frac{ y_{ 2 } – y_{ 1 } }{ x_{ 2 } – x_{ 1 } }$

Also, the equation for the line passing through the two given points (x1, y1) and (x2, y2) is given by:

y – y1 = y2y1x2x1$\frac{ y_{ 2 } – y_{ 1 } }{ x_{ 2 } – x_{ 1 } }$ ( x – x1 )

Hence, the required equation for the line which passes through (-2, 2) and (3, -5) is

y – 2  = 523(2)$\frac{ -5 – 2 }{ 3 – \left( -2 \right ) }$ ( x + 2 )

y – 2 = 75$\frac{ -7 }{ 5 }$( x + 2 )

5(y – 2) = -7 (x + 2)

5y – 10 = -7x – 14

7x + 5y + 4 = 0

Therefore, the required equation is 7x + 5y + 4 = 0

Q-8. Write the equation for the line which is at a perpendicular distance of 7 units from the point (0, 0) and the perpendicular makes an angle of 45$45^{\circ}$ along the positive x- axis.

Solution.

Let, k be the length of the normal from the origin to a line and Θ$\Theta$ be the angle of the normal to a line in the positive direction of the x- axis, then

The equation of the line is:

acosΘ+bsinΘ=k$a cos\Theta + b sin\Theta = k$

Since, k = 5 units and Θ=45$\Theta = 45^{\circ}$

Therefore,

The equation of the line satisfying the conditions is given by:

acos45+bsin45=7$a cos 45^{\circ} + b sin 45^{\circ} = 7$ a12+b12=7$a \frac{ 1 }{ \sqrt{ 2 } } + b \frac{ 1 }{ \sqrt{ 2 } } = 7$

i.e., a+b=72$a + b = 7\sqrt{ 2 }$

i.e., a+b72=0$a + b – 7\sqrt{ 2 } = 0$

Q-9. Consider a triangle ABC whose vertices are A (3, 2), B (-3, 4) and C (5, 6). Write the equation for the median through one of the vertex of the triangle say, C.

Solution.

As per the data given in the question,

The vertices of the Δ$\Delta$ABC are A( 3, 1 ), B( -3, 4 ) and C( 5, 6 ).

Let us assume that, CM be the median of the triangle through vertex C.

It means, M is the mid- point of the side AB.

Now, by mid- point theorem, the required coordinates of the point M is given by:

(332,2+42)$\left ( \frac{ 3 – 3 }{ 2 }, \frac{ 2 + 4 }{ 2 } \right )$ = ( 0, 3 )

We know that, the equation for the line passing through the two given points (a1, b1) and (a2, b2) is given by:

b – b1 = b2b1a2a1$\frac{ b_{ 2 } – b_{ 1 } }{ a_{ 2 } – a_{ 1 } }$ ( a – a1 )

Here, (a1, b1) = (5, 6) and (a2, b2) = (0, 3)

So, by substitution these values in the equation for the line CM passing through two given points is:

(b – 6) = 3605$\frac{ 3 – 6 }{ 0 – 5 }$( a – 5 )

(b – 6) = 35$\frac{ 3 }{ 5 }$( a – 5 )

5 (b – 6) = 3 (a – 5)

5b – 30 = 3a – 15

3a – 5b + 15 = 0

Therefore, the equation of the median through vertex R is 3a – 5b + 15 = 0

Q-10. Get the equation for the line which is passing through (-4, 5) and which is perpendicular to the line passing through two points (3, 6) and (-4, 7).

Solution.

We know that,

m=y2y1x2x1$m = \frac{ y_{ 2 } – y_{ 1 } }{ x_{ 2 } – x_{ 1 } }$

So, the slope of the line joining the two given points, namely A (3, 6) and B (-4, 7), i.e.,

m = 7643=17$\frac{ 7 – 6 }{ -4 – 3 } = \frac{ 1 }{ -7 }$

Now,

It is known that the two non- vertical lines are perpendicular to each other if and only if each of their slopes is a negative reciprocal of each other.

Hence, the slope of the line which is perpendicular to the line through the given points (3, 6) and (-4, 7) is:

1m=117=7$-\frac{ 1 }{ m } = -\frac{ 1 }{ \frac{ -1 }{ 7 } } = 7$

Hence, the equation for the line passing through the point (-4, 6), having slope 7 is given by:

(y – 6) = 7 (x + 4)

y – 5 = 7x + 28

i.e., 7x – y + 33 = 0

Therefore, the required equation for the line is 7x – y + 33 = 0

Q-11. Consider a line which is perpendicular to the line segment joining the two points (2, 0) and (3, 4) dividing it in 1 : p ratio. Find the equation for such a line.

Solution.

By having the section formula, we know that

The coordinates of the point dividing the line segment joining the two points (2, 0) and (3, 4) in the ratio 1 : p which is given by:

(p(2)+2(3)1+p,p(0)+2(4)1+p)=(2p+6p+1,8p+1)$\left ( \frac{ p \left ( 2 \right ) + 2 \left ( 3 \right )}{ 1 + p }\; , \; \frac{ p \left ( 0 \right ) + 2 \left ( 4 \right )}{ 1 + p } \right ) = \left ( \frac{ 2p + 6 }{ p + 1 } , \frac{ 8 }{ p + 1 }\right )$

Now, the slope for the line joining the two given points (2, 0) and (3, 4) is given by:

m = 4032=41$\frac{ 4 – 0 }{ 3 – 2 } = \frac{ 4 }{ 1 }$ = 4

It is known that the two non- vertical lines are perpendicular to each other if and only if each of their slopes is a negative reciprocal of each other.

Hence, The slope of the line which is perpendicular to the line through the given points (2, 0) and (3, 4 ) is:

1m=14$-\frac{ 1 }{ m } = -\frac{ 1 }{ 4 }$

Hence, the equation for the line passing through the point (2p+6p+1,8p+1)$\left ( \frac{ 2p + 6 }{ p + 1 } , \frac{ 8 }{ p + 1 }\right )$, having slope 14$-\frac{ 1 }{ 4 }$ is given by:

$\Rightarrow$   (y8p+1)=14(x2p+6p+1)$\left (y\; -\; \frac{ 8 }{ p + 1 } \right ) = -\frac{ 1 }{ 4 } \left ( x \;-\; \frac{ 2p + 6 }{ p + 1 } \right )\\$

$\\\Rightarrow$   4(y8p+1)=(x2p+6p+1)$4\left (y\; – \;\frac{ 8 }{ p + 1 } \right ) = -\left ( x\; – \;\frac{ 2p + 6 }{ p + 1 } \right )$

$\Rightarrow$   4(p+1)y32=x(p+1)+(2p+6)$4\left (p + 1 \right )y – 32 = -x\left ( p + 1 \right ) + \left ( 2p + 6 \right )$

$\Rightarrow$   x(p+1)4(p+1)y=(2p+632)$x\left (p + 1 \right) – 4\left (p + 1 \right)y = -\left( 2p + 6 – 32 \right )$

$\Rightarrow$   x(p+1)4(p+1)y=262p$x\left (p + 1 \right) – 4\left (p + 1 \right)y = 26 – 2p$

Therefore, the equation is x(p+1)4(p+1)y=262p$\Rightarrow x\left (p + 1 \right) – 4\left (p + 1 \right)y = 26 – 2p$.

Q-12. What will be the equation for the line which cuts off intercepts equally on the co-ordinate axes and which passes through the point (3, 4).

Solution.

It is known that, the equation of any line in the intercept form is:

xm+yn=1$\frac{ x }{ m } + \frac{ y }{ n } = 1$                    …………(i)

Where, m and n are intercepts on the x- axis and y- axis, respectively.

From the data given in the question it is known that, the line cuts off an equal intercepts on the co-ordinate axes, which clearly means that m = n.

So, from equation (i), we have

xm+ym=1$\frac{ x }{ m } + \frac{ y }{ m } = 1$

x+y=m$\Rightarrow x + y = m$                                     …………..(ii)

Thus, the line passes through the point (3, 4), equation ( ii ) reduces to

3 + 4 = m

So, m = 7

On substituting the value of m in equation (ii), we will get

x + y = 7

Hence, the equation of the line required is x + y = 7.

Q-13. A line is passing through (3, 3) and making intercepts on the coordinate axes. The sum of the two intercepts is 12. Find the equation of the line.

Solution.

It is known that, the equation of any line in the intercept form is:

xm+yn=1$\frac{ x }{ m } + \frac{ y }{ n } = 1$                    …………(i)

Where, m and n are intercepts on the x- axis and y- axis, respectively.

From the data given in the question it is known that,

m + n = 12

So, m = 12 – n

Substituting the value of m in equation (i), we will get

x12n+yn=1$\frac{ x }{ 12 – n } + \frac{ y }{ n } = 1$                    …………(ii)

Since, the line is passing through the point (3, 3). Hence, from equation (ii), we will get

312n+3n=1$\frac{ 3 }{ 12 – n } + \frac{ 3 }{ n } = 1\\$

$\Rightarrow$   3(n)+3(12n)n(12n)=1$\frac{ 3( n ) + 3 ( 12 – n )}{ n ( 12 – n )} = 1$

$\\\Rightarrow$   3n+363n12nn2=1$\frac{ 3n + 36 – 3n }{ 12n – n^{ 2 }} = 1$

$\Rightarrow$   3612nn2=1$\frac{ 36 }{ 12n – n^{ 2 }} = 1$

$\Rightarrow$   36=12nn2$36 = 12n – n^{ 2 }$

n2$\Rightarrow n^{2}$   12n+36=0$– 12n + 36 = 0$

$\Rightarrow$   n26n6n+36=0$n^{ 2 } – 6n – 6n + 36 = 0$

$\Rightarrow$   n(n6)6(n6)$n\left( n – 6 \right ) – 6\left( n – 6 \right )$

(n – 6) (n – 6) = 0

So, n = 6 or n = 6

Now,

As n = 6, so m = 6

Hence, the required equation of the line is given by:

x6+y6=1$\frac{ x }{ 6 } + \frac{ y }{ 6 } = 1\\$

Therefore, Equation is x + y = 6.

Q-14. Write the equation for the line which is passing through (0, 3) and makes an angle

π6$\frac{ \pi }{ 6 }$ along the positive direction of x- axis. Also, find the equation for the line which is parallel to it and which crosses the y-axis at the distance of 3 units to the negative direction of y- axis from (0, 0).

Solution.

As the line is making an angle π6$\frac{ \pi }{ 6 }$ in the positive direction of x- axis,

So the slope of the line is m = tanπ6$\frac{ \pi }{ 6 }$ = tan30$30^{\circ}$ = 13$\frac{ 1 }{ \sqrt{ 3 }}$

Thus, the equation of the line which is passing through (0, 3) and having slope 13$\frac{ 1 }{ \sqrt{ 3 }}$ is

(y – 3) = 13$\frac{ 1 }{ \sqrt{ 3 }}$ ( x –  0 )

3$\sqrt{ 3 }$( y – 3 ) = x

i.e., x – 3$\sqrt{ 3 }$y + 33$\sqrt{ 3 }$ = 0

Hence, the slope of the line parallel to x – 3$\sqrt{ 3 }$y + 33$\sqrt{ 3 }$ = 0 is 13$\frac{ 1 }{ \sqrt{ 3 }}$ to the line.

In the question, it is given that the line parallel to x – 3$\sqrt{ 3 }$y + 33$\sqrt{ 3 }$ = 0 crosses y- axis at a distance 3 units to the negative direction of y- axis from the origin, i.e., it passes through point ( 0, -3 ).

Therefore, the equation of the line which is passing through ( 0, -3 ) and having slope 13$\frac{ 1 }{ \sqrt{ 3 }}$ is:

y – ( -3 ) = 13$\frac{ 1 }{ \sqrt{ 3 }}$ ( x – 0 )

y + 3 = 13$\frac{ 1 }{ \sqrt{ 3 }}$ ( x – 0 )

3$\sqrt{ 3 }$( y + 3 ) = x

i.e., x – 3$\sqrt{ 3 }$y – 33$\sqrt{ 3 }$ = 0

Hence, the required equation of the line = x – 3$\sqrt{ 3 }$y – 33$\sqrt{ 3 }$ = 0.

Q-15. What will be the equation of the line perpendicular to the line from the origin meets at       (-3, 10)?

Solution.

The slope of the line which joins (0, 0) and the point (-3, 10) is

m  = 10030=103$\frac{ 10 – 0 }{ -3 – 0 } = -\frac{ 10 }{ 3 }$

So, accordingly, the slope of the line which is perpendicular to the line joining (0, 0) and the point (-3, 10) is-

m1 = –1m=1103=310$\frac{ 1 }{ m } = -\frac{ 1 }{ \frac{ -10 }{ 3 }} = { 3 } { 10 }$

Thus, the equation of the line which passes from (-3, 10) and having slope m1 is given by

(y – 10) = 310${ 3 } { 10 }$( x + 3 )

10( y – 10 ) = 3 ( x + 3 )

10y – 100 = 3x + 9

i.e., 30x – 10 y + 91 = 0

Hence, the required equation of the line is 30x – 10 y + 91 = 0

Q-16. Consider a copper rod having length L (in centimeters) which is a linear function of the Celsius temperature which is C. Express L in terms of C, for an experiment. Assume that, L = 127.952 when C = 24, and L = 128.987 when C = 114.

Solution.

As per the data given in the question, we have

If C = 24, then the value of L = 127.952, whereas if C = 114, then the value of L = 128.987.

So, the points are (24, 127.952) and (114, 128.987) which satisfies the linear relation between L and C.

Consider L along the y- axis and C along the x- axis and also, we have two points i.e., (24, 127.952) and (114, 128.987) in XY plane.

Hence, the linear relation of the line between C and L which is passing through (24, 127.952) and (114, 128.987) is given by:

(L – 127.952) = 128.987127.95211424$\frac{ 128.987 – 127.952 }{ 114 – 24 }$( C – 24 )

L – 127.952 = 1.03590$\frac{ 1.035 }{ 90 }$(C – 24)

i.e., L = 1.03590$\frac{ 1.035 }{ 90 }$( C – 24 ) + 127.952

Hence, the linear relation between C and L required = L = 1.03590$\frac{ 1.035 }{ 90 }$( C – 24 ) + 127.952

Q-17. A milk store owner observed that, he can sell at least 1000 liters of milk every week at a price of Rs. 15/liter and also, 1240 liters of milk every week at a price of Rs. 17/liter. Consider a linear relation between the demand and estimated selling price, find how many liters would he will sell every week at a cost of Rs.18/liters?

Solution.

It is given in the question that there is a linear relation between the selling price and the demand.

Consider demand along the y- axis and the estimated selling price per liter along the x- axis. (15, 1000) and (17, 1240) are the two given points along the XY plane which satisfies the linear relation between the selling price and the demand.

Hence, the linear relation of the line between S (Selling price) and D (demand) which is passing through (15, 1000) and (17, 1240) is given by:

y – 1000 = 124010001715=2402=120$\frac{ 1240 – 1000 }{ 17 – 15 } = \frac{ 240 }{ 2 } = 120$( x – 15 )

y – 1000 = 2402$\frac{ 240 }{ 2 }$( x – 15 )

y – 1000 = 120 (x – 15)

i.e., y = 120 (x – 15) + 1000

If, x = Rs. 18/liters

y = 120(18 – 15) + 1000

y = 120 (3) + 1000

y = 1360

Hence, the milk store owner will sell 1360 liters of milk weekly at the rate of Rs. 18/liter.

Q-18. S (m, n) be the mid-point of the line- segment between the axes. Prove that the equation of such line is-

xm+yn=2$\frac{ x }{ m } + \frac{ y }{ n } = 2$

Solution.

Let, PR be the line segment between the axes and let, S (m, n) be its mid-point.

Thus, the co-ordinates of P and R be (0, y) and (x, 0), respectively.

As, point S (m, n) is the mid-point of the line PR.

So, (0+x2,y+02)=(m,n)$\left( \frac{ 0 + x }{ 2 },\; \frac{ y + 0 }{ 2 } \right ) = \left( m, n \right )$

(x2,y2)=(m,n)$\Rightarrow \left( \frac{ x }{ 2 },\; \frac{ y }{ 2 } \right ) = \left( m, n \right )$ x2=mandy2=n$\Rightarrow \frac{ x }{ 2 } = m \; and \; \frac{ y }{ 2 } = n$

x = 2m and y = 2n

Hence, the coordinates of P and R are (0, 2n) and (2m, 0).

Therefore, the equation of the line which is passing through the points (0, 2n) and (2m, 0) is given by:

(y – 2n) = 02n2m0$\frac{ 0 – 2n }{ 2m – 0 }$ ( x – 0 )

(y – 2n) = 2n2m$\frac{ -2n }{ 2m }$( x – 0 )

(y – 2n) = nm$\frac{ -n }{ m }$( x )

m (y – 2n) = -n( x )

my – 2mn = -nx

i.e., nx + my – 2mn = 0

By dividing both side with mn, we will get

nxmn+mymn2mnmn=0$\frac{ nx }{ mn } + \frac{ my }{ mn } – \frac{ 2mn }{ mn } = 0$ xm+yn2=0$\frac{ x }{ m } + \frac{ y }{ n } – 2 = 0$ xm+yn=2$\frac{ x }{ m } + \frac{ y }{ n } = 2$

Hence, the required equation of the line is xm+yn=2$\frac{ x }{ m } + \frac{ y }{ n } = 2$.

Q-19. The point R (m, n) divides the line segment PQ in the ratio of 2: 3 between both the axes. What will be the equation of the line?

Solution.

As per the data given in the question, we know that PQ is the line segment which is divided in the ratio of 2 : 3 by the point R (m, n).

Let, the coordinates of P and Q is (x, 0) and (0, y), respectively.

As we know, PQ is the line segment which is divided in the ratio of 2 : 3 by the point R( m, n ).

By using the section formula,

( m, n ) = 2×0+3×x2+3,2×y+3×02+3$\frac{ 2 \times 0 + 3 \times x }{ 2 + 3 },\; \frac{ 2 \times y + 3 \times 0 }{ 2 + 3 }$

(m,n)=(3x5,2y5)$( m,\;n ) = \left(\frac{3x}{5}, \; \frac{2y}{5}\right)$

$\\\Rightarrow$ m=3x5andn=2y5$m = \frac{ 3x }{ 5 } \; and \; n = \frac{ 2y }{ 5 }$

$\Rightarrow$    x=5m3andy=5n2$x = \frac{ 5m }{ 3 } \; and \; y = \frac{ 5n }{ 2 }$

Thus, the coordinates of P and Q are ( 5m3$\frac{ 5m }{ 3 }$, 0 ) and ( 0, 5n2$\frac{ 5n }{ 2 }$ ).

Then,

The equation for the line PQ which is passing through points (5m3,0)and(0,5n2)$\left ( \frac{ 5m }{ 3 }, 0 \right ) \;and\; \left ( 0,\frac{ 5n }{ 2 } \right )$ is given by:

(y – 0) = (5n2005m3)(x5m3)$\left( \frac{ \frac{ 5n }{ 2 } – 0 }{ 0 – \frac{ 5m }{ 3 } } \right ) ( x – \frac{ 5m }{ 3 })$

y = (5n×35n×2)(x5m3)$\left( \frac{ 5n \times 3 }{ -5n \times 2 } \right ) \left( x – \frac{ 5m }{ 3 } \right )$

y = 3n2m(x5m3)$\frac{ -3n }{ 2m } \left( x – \frac{ 5m }{ 3 } \right )$

$\Rightarrow$    2m×y=3n×3x5m3$2m \times y = -3n \times \frac{ 3x – 5m }{ 3 }$

$\Rightarrow$    2m×y×3=3n×3x+3n×5m$2m \times y \times 3 = -3n \times 3x + 3n \times 5m$

6my = -9nx + 15mn

9nx + 6my – 15mn = 0

3(3nx + 2my – 5 mn) = 0

i.e., 3nx + 2my – 5mn = 0

Hence, the equation of the line PQ required is given by 3nx + 2my – 5mn = 0.

Q-20. Show that the three points (4, 1), (-3, -3) and (11, 5) are collinear by using the concept of the equation of a line.

Solution.

If we will prove that the line which passes through the points (4, 1) and (-3, -3) passes through the point (11, 5) also, then we can prove that the three points (4, 1), (-3, -3) and (11, 5) are collinear.

Now, the required equation of the line passing through (4, 1) and (-3, -3) will be given by:

(y – 1) = 3134$\frac{ -3 – 1 }{ -3 – 4 }$( x – 4 )

y – 1 = 47$\frac{ -4 }{ -7 }$( x – 4 )

7y – 7 = 4 (x – 4)

7y = 4x – 16 + 7

4x – 7y = 9

Since, the line passes through the point (11, 5), so, x = 11 and y = 5

Now,

LHS = 4 × 11 – 7 × 5 = 44 – 35 = 9 = RHS

Hence, the line which is passing through the points (4, 1) and (-3, -3) also passes through the other point (11, 5). Hence, (4, 1), (-3, -3) and (11, 5) are collinear.