The perpendicular distance from (1,2) to the straight line 12x+5y=7 is

$\frac{15}{13}$

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$\frac{12}{13}$

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$\frac{5}{13}$

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$\frac{7}{13}$

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Solution

The correct option is A
$\frac{15}{13}$

Perpendicular distance from (x1,y1) to the line ax+by+c=0 is $\frac{| a x_{1} + b y_{1} + c |}{\sqrt{(a)^{2} + b^{2}}}$

∴ perpendicular distance from (1,2) to the line 12x+5y=7 is = $\frac{| 12 + 10 - 7 |}{\sqrt{(12)^{2} + 5^{2}}} = \frac{15}{13}$

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The perpendicular distance from (1,2) to the straight line 12x+5y=7 is

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The perpendicular distance from (1,2) to the straight line 12x+5y=7 is

The correct option is A 1513 Perpendicular distance from (x1,y1) to the line ax+by+c=0 is |ax1+by1+c|√(a)2+b2 ∴ perpendicular distance from (1,2) to the line 12x+5y=7 is = |12+10−7|√(12)2+52=1513

The correct option is A
$\frac{15}{13}$

Perpendicular distance from (x1,y1) to the line ax+by+c=0 is $\frac{| a x_{1} + b y_{1} + c |}{\sqrt{(a)^{2} + b^{2}}}$

∴ perpendicular distance from (1,2) to the line 12x+5y=7 is = $\frac{| 12 + 10 - 7 |}{\sqrt{(12)^{2} + 5^{2}}} = \frac{15}{13}$