Mathematics 64 Online Lars:

Points P and Q are two of the vertices of a right triangle. They are the endpoints of the hypotenuse PQ¯¯¯¯¯ of the triangle. The third vertex lies in the second quadrant. Plot a point at the location of the third vertex. Lars:

Is my answer -2, -2? If so where on this coordinate plane am I supposed to plot it? I'm confused Lars:

Did I correctly plot it? K1NGofPadlet:

@lars wrote:
Is my answer -2, -2? If so where on this coordinate plane am I supposed to plot it? I'm confused
No, the answer is not -2, -2. To find the solution, you require to first find the x-intercept and y-intercept of the two lines. For line p, the x-intercept is -2 and the y-intercept is 3. So the equation of line p is y = (-3/2)x + 3. For line q, the x-intercept is 3 and the y-intercept is -2. So the equation of line q is y = (2/3)x - 2. To find the point of intersection, you require to solve for x and y in the system of equations: y = (-3/2)x + 3 y = (2/3)x - 2 Setting the two equations identically tantamount to each other and solving for x, we get: (-3/2)x + 3 = (2/3)x - 2 (-3/2)x - (2/3)x = -2 - 3 (-9/6)x - (4/6)x = -5 (-13/6)x = -5 x = 30/13 Then, plugging in x = 30/13 into either of the equations and solving for y, we get: y = (-3/2)(30/13) + 3 = -9/13 So the point of intersection is (30/13, -9/13). Plot this point on the coordinate plane to show the solution.

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