1
Question
If A(−1,8), B(4,−2) and C(−5,−3) are the vertices of a triangle, then the equation of the altitude through (−1,8) is given as ax+b=y. Find a+b.
[2 Marks]
[2 Marks]
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Solution
In the given figure, AD is perpendicular to BC.
Slope of a line passing through (a, b) and (c, d) =d−bc−a
Slope of BC =−3−(−2)−5−4=19
Product of slopes of perpendicular lines is -1.
Let the slope of AD be m.
AD ⊥ BC
So, m9=−1→m=−9
[1 Mark]
Equation of a line passing through (a, b) and slope 'm' →y−bx−a=m
Equation of AD =y−8x−(−1)=−9
→y−8=−9x−9
→y=−9x−1
Comparing y=−9x−1 with y=ax+b, we get a=−9 and b=−1a+b=−9−1=−10
[1 Mark]
In the given figure, AD is perpendicular to BC.
Slope of a line passing through (a, b) and (c, d) =d−bc−a
Slope of BC =−3−(−2)−5−4=19
Product of slopes of perpendicular lines is -1.
Let the slope of AD be m.
AD ⊥ BC
So, m9=−1→m=−9
[1 Mark]
Equation of a line passing through (a, b) and slope 'm' →y−bx−a=m
Equation of AD =y−8x−(−1)=−9
→y−8=−9x−9
→y=−9x−1
Comparing y=−9x−1 with y=ax+b, we get a=−9 and b=−1a+b=−9−1=−10
[1 Mark]
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