The coordinates of the foot of the perpendicu | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the foot of the perpendicular be $(x, y)$.Using the formula for the foot of the perpendicular from $(x_1, y_1)$ to $ax + by + c = 0$:$\frac{x

Vedclass · Accepted Answer

$\left(\frac{4}{5}, \frac{13}{5}\right)$

Vedclass · Answer

(A) Let the foot of the perpendicular be $(x, y)$.Using the formula for the foot of the perpendicular from $(x_1, y_1)$ to $ax + by + c = 0$:$\frac{x - x_1}{a} = \frac{y - y_1}{b} = -\frac{ax_1 + by_1 + c}{a^2 + b^2}$Substituting the values $(x_1, y_1) = (1, 2)$ and $x - 3y + 7 = 0$:$\frac{x - 1}{1} = \frac{y - 2}{-3} = -\frac{1 - 3(2) + 7}{1^2 + (-3)^2}$$\frac{x - 1}{1} = \frac{y - 2}{-3} = -\frac{1 - 6 + 7}{1 + 9} = -\frac{2}{10} = -\frac{1}{5}$Now,solve for $x$ and $y$:$x - 1 = -\frac{1}{5} \Rightarrow x = 1 - \frac{1}{5} = \frac{4}{5}$$y - 2 = -3 	imes \left(-\frac{1}{5}ight) = \frac{3}{5} \Rightarrow y = 2 + \frac{3}{5} = \frac{13}{5}$Thus,the foot of the perpendicular is $\left(\frac{4}{5}, \frac{13}{5}ight)$.

The coordinates of the foot of the perpendicular from the point $(1, 2)$ on the line $x - 3y + 7 = 0$ are

Similar Questions

If the image of $\left(\frac{-7}{5}, \frac{-6}{5}\right)$ in a line is $(1, 2)$,then the equation of the line is

Let the triangle $PQR$ be the image of the triangle with vertices $(1,3), (3,1)$ and $(2,4)$ in the line $x+2y=2$. If the centroid of $\triangle PQR$ is the point $(\alpha, \beta)$,then $15(\alpha-\beta)$ is equal to:

The equation of the perpendicular bisector of the line segment joining the points $(7, 4)$ and $(-1, -2)$ is:

If $M$ is the foot of the perpendicular drawn from the origin to the line $x-2y+3=0$ which meets the $X$ and $Y$-axes at $A$ and $B$ respectively,then $AM=$

The image of the point $(4, -13)$ with respect to the line $5x + y + 6 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module