A. 2x + 3y = 7
-x + 2y = -7
Multiply Eq2 by 2 and add the Eqs.
2x + 3y = 7
-2x + 4y = -14
sum = 7y = -7
Y = -1
In Eq1, replace y with -1 and solve for x:
2x + 3*(-1) = 7
2x = 10
X = 5
P(5,-1)
B. P(5,-1), Q(-9,2).
D^2=(-9-5)^2 + (2-(-1))^2=196 + 9 = 205
D = 14.3
Y = mx + b
m = (2-(-1))/(-9-5) = 3/-14 = -3/14
Y = (-3/14)5 + b = -1
-15/14 + b = -1
b = -1 + 15/14 = 1/14
Y=(-3/14)x + 1/14. Slope intercept form.
14y = -3x + 1
3x + 14y = 1. Standard form.
The two straight lines 3y + 2x - 7= 0 and 2y =x
-7 intersect at the point P.(a) Find the coordinates if P.
(B) Q is the point (-9,2).Find (1) the distance PQ.
(2)the equation of PQ.
1 answer
1 answer