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The bisector of the acute angle formed between the lines 4x - 3y + 7 = 0 and 3x - 4y + 14 = 0 has the equationA. x + y + 3 = 0B. x - y - 3 = 0C. x - y + 3 =0D. 3x - y - 7 = 0 |
Answer» Correct Answer - C The equations of given straight lines , by making constant terms positive , are 4x - 3y + 7 = 0 and 3x - 4y + 14 = 0 `because 4 xx 3 + (-3) (-4) = 24 gt 0 ` i.e, `a_(1) a_(2) + b_(1)b_(2) gt 0 ` So , the bisector of the acute angle is given by `(4x - 3y + 7)/(sqrt(4^(2) + (-3)^(2))) = -(3x - 4y + 14)/(sqrt(3^(2) (-4)^(2))) implies x - y + 3 = 0 ` |
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