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9.Straight Line
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ત્રણ રેખાઓ $4x - 7y + 10 = 0; x + y=5$ અને $7x + 4y = 15$ થી રચાતા ત્રિકોણના લંબકેન્દ્રના યામો મેળવો
A
$(2, 1)$
B
$(- 1, 2)$
C
$(1, 2)$
D
$(1, - 2)$
Solution
Given lines $4 x-7 y+10=0—-(1)$
$x+y-5=0—-(2)$
$7 x+4 y-15=0—-(3)$
On solving eq ( 1 ) and ( 3 ) $7 x+4\left(\frac{4 x+10}{7}\right)-15=0$
$49 x+16 x+40-105=0$
$65 x=65 \Rightarrow x=1$
From eq (1) $4-7 y=-10$
$-7 y=-14$
$y=2$
Point of intersection is (1,2) Here Equation (1) and (3) are perpendicular So the lines formed right angle triangle Hence the orthocentre of Right angle triangle is the point at which $90^{0}$ angle is formed Hence (1,2) is orthocentre
Standard 11
Mathematics
