Solve the linear equation of point P (- 5,4) which is surrounded by two coordinate axes and has a triangle area of 5

Solve the linear equation of point P (- 5,4) which is surrounded by two coordinate axes and has a triangle area of 5

Let: the slope of the straight line be K, then the equation of the straight line is: y-4 = K (x + 5) let x = 0 and y = 0 seek y and y = 0 seek x respectively, then the intersection coordinates of the straight line and Y axis and X axis are: (0,5k + 4), ((5K + 4) / K, 0) because the area of the triangle formed by the straight line and two coordinate axes is 5, so: (1 / 2) · I (5K + 4) · [- (5K + 4) / k] I = 5