Answer:
[tex]y = \frac{3}{4} x+ \frac{7}{4}[/tex]
Step-by-step explanation:
The slope of straight line PR where P(2,3) and R(5,-1) are two vertices of triangle PQR will be = [tex]\frac{3-(-1)}{2-5} =-\frac{4}{3}[/tex]
Therefore, the slope of the altitude passing through Q(-1,1) will be [tex]\frac{3}{4}[/tex] {Since, the product of slopes of two perpendicular straight line is -1}
So, equation of the altitude is [tex]y=\frac{3}{4} x + c[/tex] where c is a constant.
Now, putting x = -1 and y = 1 in the above equation we get
[tex]1 = -\frac{3}{4} + c[/tex]
⇒ [tex]c=\frac{7}{4}[/tex]
Therefore, the equation of the altitude is [tex]y = \frac{3}{4} x+ \frac{7}{4}[/tex] (Answer)
