Question

through: (2,5), perp. to y =- 2/3x +3

Answer 1

Find the equation of line passes through point (2, 5) and perpendicular to line having equation y =- 2/3x +3

Answer:

The equation of line passes through point (2, 5) and perpendicular to line having equation y =- 2/3x +3 in slope intercept form is $y = \frac{3}{2}x + 2$

Solution:

Given that line passes through (2, 5) and perpendicular to line having equation $y = \frac{-2}{3}x + 3$

Let us first find slope of original line

The slope intercept form of line is given as:

y = mx + c ------ eqn 1

Where "m" is the slope of line and "c" is the y - intercept

On comparing the slope intercept form y = mx + c and given equation of line $y = \frac{-2}{3}x + 3$ we get

$m = \frac{-2}{3}$

Thus slope of given line is $m = \frac{-2}{3}$

We know that product of slopes of given line and slope of line perpendicular to given line is always -1

Slope of given line x slope of line perpendicular to it = -1

$\begin{array}{l}{\frac{-2}{3} \times \text { slope of line perpendicular to it }=-1} \\\\ {\text { slope of line perpendicular to it }=\frac{3}{2}}\end{array}$

Let us find equation of line with slope $m = \frac{3}{2}$ and passes through (2, 5)

Substitute $m = \frac{3}{2}$ and (x, y) = (2, 5)

$5 = \frac{3}{2} \times 2 + c\\\\5 = 3 + c\\\\c = 2$

Thus the required equation of line is:

Substitute $m = \frac{3}{2}$ and c = 2 in eqn 1

$y = \frac{3}{2}x + 2$

Thus the equation of line perpendicular to given line is found out