One vertex of square is (−4,5) and equation of one diagonal is 7x−y+8=0 Diagonal of a square are perpendicular and bisect each other Let the equation of the other diagonal be y=mx+c where m is the slope of the line and c is the y-intercept. Since this line passes through (−4,5) ∴5=−4m+c‌. ‌ Since this line is at right angle to the line 7x−y+8=0 or y=7x+8, having slope =7, ∴7×m=−1 or m=‌
−1
7
Putting this value of m in equation (i) we get 5=−4×(‌