Condition for Two Lines to Be Perpendicular

Q. The line passing through the extremity A of the major axis and extremity B of the minor axis of the ellipse $x^2+9y^2=9$, meets its auxiliary circle at the point M. Then the area of the triangle with vertices at A, M and the origin ‘O’ is

1. 
2. 
3. 
4. 

Q.

In the triangle ABC with vertices A(2, 3), B (4, -1) and C(1, 2), find the equation and length of altitude from the vertex A

Q.

Find the equation of a line passing through (2, 3) and inclined at an angle of ${135}^{\circ }$ with the positive direction of x-axis

1. x + y - 5 = 0

2. x + y + 1 = 0

3. x + y - 1 = 0

4. x + y + 5 = 0

Q. Find equation of the line perpendicular to the line x – 7 y + 5 = 0 and having x intercept 3.

Q.

If the area of the parallelogram whose sides are x + 2y + 3 = 0, 3x + 4y - 5 = 0, 2x+4y+5=0 and 3x + 4y - 10 = 0 is 'a' sq. unit. Find the value of 4a

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Q.

Reduce the equation $3x-2y+4=0$ to intercepts form and find the length of the segment intercepted between the axes.

Q. If the straight line, $2x-3y+17=0$ is perpendicular to the line passing through the points $(7,17)$ and $(15,\beta )$ then $\beta$ equals:

1. $-5$
2. $\frac{35}{3}$
3. $5$
4. $-\frac{35}{3}$

Q. A line L is perpendicular to the line 5x-y=1 and the area of the triangle formed by the line L and coordinate axes is 5. The equation of the line L is

1. $x+5y=5$
2. $x+5y=\pm 5\sqrt{2}$
3. $x-5y=5$
4. $x-5y=5\sqrt{2}$

Q. Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right angled triangle.

Q. If $l_1,m_1,n_1$ and $l_2,m_2,n_2$ are the direction cosines of two perpendicular lines, then the direction cosine of the line which is perpendicular to both the lines, will be

1. $(m_1{n}_2-m_2{n}_1),(n_1{l}_2-l_1{n}_2),(l_1{m}_2-l_2{m}_1)$
2. $(l_1{l}_2-m_2{m}_1),(m_1{m}_2-n_1{n}_2),(n_1{n}_2-l_2{l}_1)$
3. $\frac{1}{\sqrt{l_1^2+m_1^2+n_1^2}},\frac{1}{\sqrt{l_2^2+m_2^2+n_2^2}},\frac{1}{\sqrt{3}}$
4. $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$

Q.

Q(h, k) is the foot of the perpendicular of P(3, 6) on the line x - 2y + 4 =0.

If the slope of PQ is m, find $m^2$.

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Q. If the lines joining the points $(K,2),(3,6)$ and $(2,3),(K,7)$ are perpendicular to each other, then the possible value of $K$ is

1. $\frac{5+\sqrt{65}}{2}$
2. $\frac{5-\sqrt{65}}{2}$
3. $\frac{5-\sqrt{65}}{4}$
4. $\frac{5+\sqrt{65}}{4}$

Q. A person standing at the junction (crossing) of two straight paths represented by the equations $2x-3y+4$ $=$ $0$ and $3x+4y-5=$ $0$ wants to reach the path whose equation is $6x-7y+8=$ $0$ in the least time. Find equation of the path that he should follow.

Q.

Q(h, k) is the foot of the perpendicular of P(3, 6) on the line x - 2y + 4 =0.

If the slope of PQ is m, find $m^2$.

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Q. If the foot of the perpendicular drawn from the point $(1,0,3)$ on a line passing through $(\alpha ,7,1)$ is $(\frac{5}{3},\frac{7}{3},\frac{17}{3})$, then $\alpha$ is equal to

Q. A person standing at the junction (crossing) of two straight paths represented by the equations 2 x – 3 y + 4 = 0 and 3 x + 4 y – 5 = 0 wants to reach the path whose equation is 6 x – 7 y + 8 = 0 in the least time. Find equation of the path that he should follow.

Q. If the $p^{th},q^{th}$ and $r^{th}$ terms of a G.P. are $a,b$ and $c$, respectively. Prove that $a^{q-r}{b}^{r-p}{c}^{p-q}=1.$

Q. If the line through the points $(-2,6)$ and $(4,8)$ is perpendicular to the line through the points $(8,12)$ and $(x,24)$, then the value of $x$ is

Q. If the two lines $(a+b)x-4y+5=0$ and $x+(a-b)y+10=0$ are perpendicular to each other then

1. $a=b$
2. $2a+b=0$
3. $3a-5b=0$
4. $5a-2b=0$

Q. Let a square with side length ${}^{\prime }{p}^{\prime }$ and making an angle of $\theta$ with $x-$ axis, has one vertex at origin. If $0<\theta <\frac{\pi }{2}$, then the equation of the diagonals of the square is

1. $x(1+\tan \theta )-y(1+\tan \theta )=py(1+\tan \theta )=x(1-\tan \theta )$
2. $x(\cos \theta +\sin \theta )-y(\cos \theta +\sin \theta )=py(\cos \theta +\sin \theta )=x(\cos \theta +\sin \theta )$
3. $x(\cos \theta -\sin \theta )+y(cos\theta +\sin \theta )=py(\cos \theta -\sin \theta )=x(\cos \theta +\sin \theta )$
4. $x(1+\tan (\frac{\pi }{4}+\theta )-y(1+\tan (\frac{\pi }{4}+\theta )=py\left(\tan \right(\frac{\pi }{4}+\theta \left)\right)=x\left(\tan \right(\frac{\pi }{4}-\theta \left)\right)$