JEE Main Maths Cartesian System and Straight Lines Previous Year Questions With Solutions

JEE Main Previous year Cartesian System and Straight Lines solutions offer greater insight into all the questions and the detailed answers provided here will further help students to take their preparation to a higher level. In this article, JEE aspirants can practice previous years’ questions on Cartesian System and Straight Lines asked in IIT JEE exams. The solutions have been prepared by our experts and they have provided accurate answers to all the listed questions which students can refer to and be ready to solve questions at the time of their examinations. We have included all the important maths questions and a detailed solution for each question. About 2-3 questions are asked from this chapter in the examination.

The important concepts such as distance between two points, coordinates of the point dividing any line segment, area of a triangle, the inclination of line, the angle between two lines, equation of the horizontal line, equation of the vertical line, a point-slope form of a line, the two-point form of a line, slope-intercept form of a line, the normal form of a line, the distance of a point from a line and perpendicular distance between two lines are covered here.

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JEE Main Cartesian System and Straight Lines Past Year Questions With Solutions

Question 1: ABC is an isosceles triangle. If the coordinates of the base are B (1,3) and C (- 2, 7). Find the coordinates of vertex A.

Solution:

The vertex A (x, y) is equidistant from B and C because ABC is an isosceles triangle. Therefore, (x1)2+(y3)2=(x+2)2+(y7)26x8y+43=0{{(x-1)}^{2}}+{{(y-3)}^{2}}={{(x+2)}^{2}}+{{(y-7)}^{2}} \rightarrow \,\,6x-8y+43=0

Thus, any point lying on this line can be the vertex A except the midpoint (12,5)\left( -\frac{1}{2},\,\,5 \right) of BC.

Question 2: If the vertices P, Q, R of a triangle PQR are rational points, which of the following points of the triangle PQR is (are) always rational point(s)?

A) Centroid

B) Incentre

C) Circumcentre

D) Orthocentre (A rational point is a point both of whose coordinates are rational numbers)

Solution:

If A=(x1,y1),B=(x2,y2),C=(x3,y3),A=({{x}_{1}},\,\,{{y}_{1}}),\,\,B=({{x}_{2}},\,\,{{y}_{2}}),\,\,C=({{x}_{3}},\,\,{{y}_{3}}), where x1,y1,{{x}_{1}},\,\,{{y}_{1}}, are rational numbers then Σx1,Σy1\Sigma {{x}_{1}},\,\,\Sigma {{y}_{1}} are also rational. So, the coordinates of the centroid (Σx13,Σy13)\left( \frac{\Sigma {{x}_{1}}}{3},\,\,\frac{\Sigma {{y}_{1}}}{3} \right) will be rational. As AB=c=(x1x2)2+(y1y2)2,cAB=c=\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}},}\,\, c may or may not be rational and it may be an irrational number of the form p\sqrt{p} Hence, the coordinates of the incentre (Σax1Σa,Σay1Σa)\left( \frac{\Sigma a{{x}_{1}}}{\Sigma a},\,\,\frac{\Sigma a{{y}_{1}}}{\Sigma a} \right) may or may not be rational. If (α, β) be the circumcentre or orthocentre, a and b are found by solving two linear equations in α, β with rational coefficients. So α, β must be rational numbers.

Question 3: The new coordinates of a point (4, 5), when the origin is shifted to the point (1,-2) are

A) (5, 3)

B) (3, 5)

C) (3, 7)

D) None of these

Solution:

We know that if the origin is shifted to (h, k), then new coordinates (x, y) becomes (x – h, y – k). Therefore, the new coordinate of (4, 5) with respect to new origin (1, – 2) is (3, 7).

Question 4: If P = (1,0), Q = (-1, 0) and R = (2, 0) are three given points, then the locus of a point S satisfying the relation SQ2+SR2=2SP2S{{Q}^{2}}+S{{R}^{2}}=2S{{P}^{2}} is

A) A straight line parallel to the x-axis

B) A circle through the origin

C) A circle with centre at the origin

D) A straight line parallel to the y-axis

Solution:

Let S (x, y), then (x+1)2+y2+(x2)2+y2=2[(x1)2+y2]2x+1+44x=4x+2x=32{{(x+1)}^{2}}+{{y}^{2}}+{{(x-2)}^{2}}+{{y}^{2}}=2\,[{{(x-1)}^{2}}+{{y}^{2}}] \\ \Rightarrow \,\,2x+1+4-4x=-4x+2\,\,\Rightarrow \,\,x=-\frac{3}{2}

Hence it is a straight line parallel to the y-axis.

Question 5: The coordinates of the points O, A and B are (0, 0), (0, 4) and (6, 0), respectively. If a points P moves such that the area of ΔPOA is always twice the area of ΔPOB, then find the equation to both parts of the locus of P.

Solution:

The three given points are O(0,0),A(0,4)O\,\,(0,\,\,0),\,\,A(0,\,\,4) and B(6,0)B\,(6,\,\,0) and let P (x, y) be the moving point.

Area of ΔPOA = 2 * Area of ΔPOB

12×4×x=±2×12×6×yx=±3y\Rightarrow \,\,\frac{1}{2}\times 4\times x=\pm \,2\times \frac{1}{2}\times 6\times y \\ x=\pm \,3y

Hence, the equation to both parts of the locus of P is (x3y)(x+3y)=0.(x-3y)\,(x+3y)=0.

Question 6: Find the Orthocentre of the triangle formed by the lines x + y = 1 and xy = 0.

Solution:

Given lines are x + y = 1 and xy = 0 when x = 0, then y = 1 when x = 1, then y = 0

∴ (0, 1) and (1, 0) are the vertices of triangle.

Clearly, the triangle is right-angled isosceles.

Orthocentre of a right-angled triangle is the same as the vertex of the right angle.

Therefore, the point of intersection of x + y = 1 and xy = 0 is (0, 0).

Question 7: Find the circumcentre of a triangle formed by the line xy + 2x + 2y + 4 = 0 and x + y + 2 = 0.

Solution:

xy + 2x + 2y + 4 = 0 …..(i) and x + y + 2 = 0 ……(ii)

From (i) and (ii), xy = 0 ⇒ x = y = 0

Vertices of triangle are (−2, 0) (0, 0) (0, −2) (In a right angled triangle circumcentre is mid point of hypotenuse)

(−1, −1) is the circumcircle.

Question 8: Find the area of the triangle formed by the points (a,b + c), (b, c + a), (c, a + b).

Solution:

Area =12ab+c1bc+a1ca+b1=12aa+b+c1ba+b+c1ca+b+c1=(a+b+c)2a11b11c11=0.=\frac{1}{2}\,\left| \,\begin{matrix} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \\ \end{matrix}\, \right|=\frac{1}{2}\,\left| \,\begin{matrix} a & a+b+c & 1 \\ b & a+b+c & 1 \\ c & a+b+c & 1 \\ \end{matrix}\, \right|\\ =\frac{(a+b+c)}{2}\,\left| \,\begin{matrix} a & 1 & 1 \\ b & 1 & 1 \\ c & 1 & 1 \\ \end{matrix}\, \right|\\ =0.

Question 9: Three points are A (6, 3), B (−3, 5), C (4, −2) and P (x, y) is a point, then the ratio of the area of ΔPBC and ΔABC is

Solution:

ΔPBCΔABC=[3(2y)+4(y5)+x(5+2)][6(5+2)3(23)+4(35)]=7x+7y1449=x+y27.\frac{\Delta \,PBC}{\Delta \,ABC}=\left| \frac{[-3(-2-y)+4(y-5)+x(5+2)]}{[6\,(5+2)-3(-2-3)+4(3-5)]} \right|\\ =\left| \frac{7x+7y-14}{49} \right|=\left| \frac{x+y-2}{7} \right|.

Question 10: If P (1, 2), Q (4, 6) R (5, 7) and S (a, b) are the vertices of a parallelogram PQRS, then find the values of a and b.

Solution:

Diagonals cut each other at middle points. Hence, a+42=1+52a=2b+62=2+72b=3.\frac{a+4}{2}=\frac{1+5}{2}\,\,\Rightarrow \,\,a=2\\ \frac{b+6}{2}=\frac{2+7}{2}\,\,\Rightarrow \,\,b=3.

Question 11: The locus of the point P (x, y) satisfying the relation (x3)2+(y1)2+(x+3)2+(y1)2=6\sqrt{{{(x-3)}^{2}}+{{(y-1)}^{2}}}+\sqrt{{{(x+3)}^{2}}+{{(y-1)}^{2}}}=6 is

Solution:

(x3)2+(y1)2+(x+3)2+(y1)2=6(x3)2+(y1)2=6(x+3)2+(y1)2\sqrt{{{(x-3)}^{2}}+{{(y-1)}^{2}}}+\sqrt{{{(x+3)}^{2}}+{{(y-1)}^{2}}}=6\\ \sqrt{{{(x-3)}^{2}}+{{(y-1)}^{2}}}=6-\sqrt{{{(x+3)}^{2}}+{{(y-1)}^{2}}}

Squaring both sides.

12x+36=12(x+3)2+(y1)212x+36=12\sqrt{{{(x+3)}^{2}}+{{(y-1)}^{2}}}

Again, squaring we get the given equation is a pair of straight lines.

Question 12: The equations to a pair of opposite sides of a parallelogram are x25x+6=0{{x}^{2}}-5x+6=0 and y26y+5=0.{{y}^{2}}-6y+5=0. Find the equations of its diagonals.

Solution:

Equation of diagonal d1{{d}_{1}} is y1=5132(x2)y1=41(x2)y=4x7y-1=\frac{5-1}{3-2}(x-2)\\ \Rightarrow y-1=\frac{4}{1}(x-2) \Rightarrow y=4x-7

Equation of diagonal d2{{d}_{2}} is y1=5123(x3)y1=4(x3)4x+y=13y-1=\frac{5-1}{2-3}(x-3)\\ \Rightarrow y-1=-4(x-3) \Rightarrow 4x+y=13

So equations are, 4x + y = 13 and y = 4x − 7.

Question 13: The area bounded by the angle bisectors of the lines x2y2+2y=1{{x}^{2}}-{{y}^{2}}+2y=1 and the line x+y=3,x+y=3, is

Solution:

The angle bisectors of the lines given by x2y2+2y=1 are x=0,y=1.{{x}^{2}}-{{y}^{2}}+2y=1 \text \ are \ x=0, y=1.

Therefore, required area =12×2×2=2.=\frac{1}{2}\times 2\times 2=2.

Question 14: The lines represented by the equation ax2+2hxy+by2+2gx+2fy+c=0a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0 will be equidistant from the origin, if

A)f2+g2=c(ba)B)f4+g4=c(bf2+ag2)C)f4g4=c(bf2ag2)D)f2+g2=af2+bg2A) {{f}^{2}}+{{g}^{2}}=c(b-a)\\ B) {{f}^{4}}+{{g}^{4}}=c(b{{f}^{2}}+a{{g}^{2}})\\ C) {{f}^{4}}-{{g}^{4}}=c(b{{f}^{2}}-a{{g}^{2}})\\ D) {{f}^{2}}+{{g}^{2}}=a{{f}^{2}}+b{{g}^{2}}

Solution:

Let the equations represented by ax2+2hxy+by2+2gx+2fy+c=0a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0 be lx+my+n=0and lx+my+n=0.lx+my+n=0 \text and \ l’x+m’y+n’=0. Then, the combined equation represented by these lines is given by (lx+my+n)(lx+my+n)=0.(lx+my+n)(l’x+m’y+n’)=0.. So, it must be similar with the given equation.

On comparing, we get

ll=a, mm=b  nn=c,  lm+ml=2h,  ln+ln=2g,mn+nm=2fll’=a,\,\ mm’=b\ \ nn’=c,\ \ lm’+ml’=2h,\ \ ln’+l’n=2g, \\ mn’+nm’=2f

According to the condition, the length of perpendiculars drawn from the origin to the lines are the same.

So,

nl2+m2=nl2+m2=(nn)2(l2+m2)(l2+m2)\frac{n}{\sqrt{{{l}^{2}}+{{m}^{2}}}}=\frac{n’}{\sqrt{l{{‘}^{2}}+m{{‘}^{2}}}}=\frac{{{(nn’)}^{2}}}{({{l}^{2}}+{{m}^{2}})(l{{‘}^{2}}+m{{‘}^{2}})}

Now on eliminating l, m, l′, m′and n, n′, we get the required condition

f4g4=c(bf2ag2).{{f}^{4}}-{{g}^{4}}=c(b{{f}^{2}}-a{{g}^{2}}).

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