Question

A line passes through (-2,-3) and is perpendicular to the line -9x+15y=4. Find it’s x-intercept and y-intercept.

Answer 1

Answer:

x-intercept and y-intercept of the given line is  $\left(-\frac{1}{5}\right)$ and $\left(\frac{1}{3}\right)$

Solution:

The equation of the line is $-9x+15y=4$

Simplifying this we get,

$\Rightarrow 9x+15y=4$

$\Rightarrow 15y =9x+4$

$\Rightarrow y=\left(\frac{9 x}{15}\right)+\left(\frac{4}{15}\right)$

The slope of the line is (9/15), so the slope of the line perpendicular to this will be $\frac{15}{9}$

Let us assume that the y intercept is b, so the equation is $y = mx + b$

Now, as that line passes through (-2,-3), hence using this point we get,

$-3=\left(\frac{15}{9}\right) \times(-2)+b$

$-3=\left(\frac{5}{3}\right) \times(-2)+b$

$-3=-\frac{10}{3}+b$

$-9=-10+3 b$

$3 b=1$

$b=\left(\frac{1}{3}\right)$

So the equation will be,  

$y=\left(\frac{15}{9}\right) \times x+\left(\frac{1}{3}\right)$

Now to find x intercept y =0, Hence,

$0=\frac{15 x}{9}+\frac{1}{3}$

$0=\frac{15 x+3}{9}$

$(15 x+3)=0$

$x=-\frac{3}{15}=-\frac{1}{5}$

To find y intercept x = 0, Hence,

$y={(\frac{15}{9}})\times0+{(\frac{1}{3}})$

$\Rightarrow y=\frac{1}{3}$

So, x intercept is $\left(-\frac{1}{5}\right)$ and y intercept is $\left(\frac{1}{3}\right)$