The equation for line c is y=13x+4. Line d is perpendicular to line c and passes through the point (–4,–4). At what point do lines c and d intersect to form a right angle?

First, we need to find the slope of line d. Since line d is perpendicular to line c, and the slope of line c is 13, the slope of line d is the negative reciprocal of 13, which is -1/13.

The equation of a line is y = mx + b, where m is the slope and b is the y-intercept. We know that line d passes through the point (-4, -4) and has a slope of -1/13, so we can substitute these values into the equation to find the y-intercept.

-4 = -1/13 * -4 + b -4 = 4/13 + b b = -4 - 4/13 b = -56/13

So, the equation of line d is y = -1/13x - 56/13.

Now, we need to find the point where lines c and d intersect. This is the point where their y values are equal, so we can set the two equations equal to each other and solve for x:

13x + 4 = -1/13x - 56/13 Multiply everything by 13 to get rid of the fractions: 169x + 52 = -x - 56 Add x and 56 to both sides: 170x + 108 = 0 Subtract 108 from both sides: 170x = -108 Divide by 170: x = -108/170 Simplify the fraction: x = -54/85

Substitute x = -54/85 into the equation for line c to find the y-coordinate of the intersection point:

y = 13(-54/85) + 4 y = -702/85 + 4 y = -702/85 + 340/85 y = -362/85

So, the lines c and d intersect at the point (-54/85, -362/85).