Foot of Perpendicular from a Point on a Line

Q.

The shaded region is represented by:

1. $4x–2y\le 3$

2. $4x–2y\le –3$

3. $4x–2y\ge 3$

4. $4x–2y\ge –3$

Q. The coordinates of the foot of perpendicular drawn from the point $(2,4)$ on the line $x+y=1$, are

1. $(\frac{1}{2},\frac{3}{2})$
2. $(-\frac{1}{2},\frac{3}{2})$
3. $(\frac{3}{2},-\frac{1}{2})$
4. $(-\frac{1}{2},-\frac{3}{2})$

Q. If the equation of the curve on reflection of the ellipse $\frac{(x-4{)}^2}{16}+\frac{(y-3{)}^2}{9}=1$ about the line $x-y-2=0$ is $16x^2+9y^2+k_{1}x-36y+k_2=0$, then $\frac{k_1+k_2}{22}$ is

Q. The area of the quadrilateral formed by the lines $4x-3y-a=0,3x-4y+a=0$, $4x-3y-3a=0$ and $3x-4y+2a=0$ is

1. $\frac{2a^2}{11}\text{sq. units}$
2. $\frac{2a^2}{9}\text{sq. units}$
3. $\frac{a^2}{7}\text{sq. units}$
4. $\frac{2a^2}{7}\text{sq. units}$

Q.

The coordinates of the image of the origin O with respect to the straight line x+y+1=0 are

1. $(-\frac{1}{2},-\frac{1}{2})$

2. $(-2,-2)$

3. $(1,1)$

4. $(-1,-1)$

Q. The equations of the perpendicular bisectors of the sides $AB$ and $AC$ of a triangle $ABC$ are $x-y+5=0$ and $x+2y=0$ respectively. If the co-ordinates of $A$ are $(1,-2)$, then the equation of $BC$ is

1. $23x+14y-40=0$
2. $14x+23y-40=0$
3. $23x-14y+40=0$
4. None of these

Q.

If PM is the perpendicular from P (2, 3) on to the line x+y=3 then the co-ordinates of M are

1. (2, 1)

2. (−1, 4)

3. (1, 2)

4. (4, −1)

Q. The line $\frac{x}{a}+\frac{y}{b}=1$ moves in such a way that $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$ where $c$ is a constant. The locus of foot of perpendicular from the origin on the given line will be

1. $x^2+y^2=a^2$
2. $x^2+y^2=b^2$
3. $x^2+y^2=c^2$
4. $x^2+y^2=(a^2-b^2{)}^2$

Q. A straight line passes through a fixed point $(h,k)$. The locus of the foot of perpendicular on it drawn from the origin is

1. $x^2+y^2=hx-ky$
2. $x^2+y^2=kx-hy$
3. $x^2+y^2-hx-ky=0$
4. $x^2+y^2-kx-hy=0$