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

Answer:x-intercept and y-intercept of the given line is  [tex]\left(-\frac{1}{5}\right)[/tex] and [tex]\left(\frac{1}{3}\right)[/tex]Solution: The equation of the line is [tex]-9x+15y=4[/tex]Simplifying this we get, [tex]\Rightarrow 9x+15y=4[/tex][tex]\Rightarrow 15y =9x+4[/tex][tex]\Rightarrow y=\left(\frac{9 x}{15}\right)+\left(\frac{4}{15}\right)[/tex] The slope of the line is (9/15), so the slope of the line perpendicular to this will be [tex]\frac{15}{9}[/tex]Let us assume that the y intercept is b, so the equation is [tex]y = mx + b[/tex]Now, as that line passes through (-2,-3), hence using this point we get, [tex]-3=\left(\frac{15}{9}\right) \times(-2)+b[/tex][tex]-3=\left(\frac{5}{3}\right) \times(-2)+b[/tex][tex]-3=-\frac{10}{3}+b[/tex][tex]-9=-10+3 b[/tex][tex]3 b=1[/tex][tex]b=\left(\frac{1}{3}\right)[/tex]So the equation will be,  [tex]y=\left(\frac{15}{9}\right) \times x+\left(\frac{1}{3}\right)[/tex]Now to find x intercept y =0, Hence, [tex]0=\frac{15 x}{9}+\frac{1}{3}[/tex][tex]0=\frac{15 x+3}{9}[/tex][tex](15 x+3)=0[/tex][tex]x=-\frac{3}{15}=-\frac{1}{5}[/tex]To find y intercept x = 0, Hence, [tex]y={(\frac{15}{9}})\times0+{(\frac{1}{3}})[/tex][tex]\Rightarrow y=\frac{1}{3}[/tex]So, x intercept is [tex]\left(-\frac{1}{5}\right)[/tex] and y intercept is [tex]\left(\frac{1}{3}\right)[/tex]