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Solve the two equations
simultaneously
2x - 3y = 7
4x + 3y = 5
6x = 12, x = 2
Substitute in 2x - 3y = 7
4 - 3y = 7; -3y = 3
y =-1
Distance of point (2,-1) from (4,3)
Distance = $\sqrt{(4 - 2)^2 + (3--1)^2}$
= $\sqrt{4 + 16}$ = $\sqrt20$
= $\sqrt{4\times 5}$ = 2$\sqrt5$
johnmulu answered the question on March 6, 2017 at 09:50
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