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  • The distance of the point of intersection of the lines 2x minus 3y plus 5 is equal to 0 and 3x plus 4y is equal to 0 from the line 5x minus 2y is equal to 0 is A 130 divided 17 square root 29 B 13 divided 7 square root 29 130 divided 7 D. None of these

The distance of the point of intersection of the lines 2x – 3y + 5 = 0 and 3x + 4y = 0 from the line 5x – 2y = 0 is
A. \frac{130}{17\sqrt{29}} 
B. \frac{13}{7\sqrt{29}}
C. \frac{130}{7}
D. None of these

Answers (1)

(a) Given lines are:
2 x-3 y+5=0
and
3 x+4 y=0
Solving these lines, we get point of intersection as \left(\frac{-20}{17}, \frac{15}{17}\right) 
therefore Distance of this point from the line 5 x-2 y=0
=\frac{\left|5 \times\left(-\frac{20}{17}\right)-2\left(\frac{15}{17}\right)\right|}{\sqrt{25+4}}=\frac{\left|\frac{-100}{17}-\frac{30}{17}\right|}{\sqrt{29}}=\frac{130}{17 \sqrt{29}}

Posted by

Satyajeet Kumar

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