JAMB - Mathematics (1997 - No. 29)
Find the distance between the point Q (4,3) and the point common to the lines 2x - y = 4 and x + y = 2
3√10
3√5
√26
√13
Explanation
2x - y .....(i)
x + y.....(ii)
from (i) y = 2x - 4
from (ii) y = -x + 2
2x - 4 = -x + 2
x = 2
y = -x + 2
= -2 + 2
= 0
\(x_1\) = 2
\(y_1\) = 0
\(x_2\) = 4
\(y_2\) = 3
Hence, dist. = \(\sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}\)
= \(\sqrt{(3 - 0)^2 + (4 - 2)^2}\)
= \(\sqrt{3^2 + 2^2}\)
= \(\sqrt{13}\)
x + y.....(ii)
from (i) y = 2x - 4
from (ii) y = -x + 2
2x - 4 = -x + 2
x = 2
y = -x + 2
= -2 + 2
= 0
\(x_1\) = 2
\(y_1\) = 0
\(x_2\) = 4
\(y_2\) = 3
Hence, dist. = \(\sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}\)
= \(\sqrt{(3 - 0)^2 + (4 - 2)^2}\)
= \(\sqrt{3^2 + 2^2}\)
= \(\sqrt{13}\)
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