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A triangle is formed by the lines whose equations are AB: x+y-5=0, BC: x+7y-7=0 and CA: 7x+y+14=0. Then |
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Answer» angle at A is acute The slopes of the LINE AB, BC and CA are -1, `-(1)/(7)` and -7, respectively, `"LET " m_(1) = -(1)/(7), m_(2) = -1, m_(3) =-7.` `thereforem_(1) gt m_(2) gtm_(3)` So, tangent fo internal angles of the triangle are `"tan" A = (3)/(4), "tan" B = (3)/(4) " and tan" C = -(24)/(7)` So, interior angles A and B are acute and interior angle C is abtuse. `therefore " Internal bisector of B "-= " Acute angle bisector at B"` `"" -= " Acute angle bisector of lines AB and BC "` `"" -= 3x+6y-16=0` External bisector of C `-= " Acute bisector of C" ` ` "" -= " Acute angle bisector of lines AC and BC"` ` "" -=8x+8y+7=0` |
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