Suppose P and Q lie on 3x+4y−4=0 and 5x−y−4=0 respectively. If the midpoint of PQ is (1,5), then the slope of the line passing through P and Q is
Let the line PQ be y−5=m(x−1)......(i) Substituting y=mx+5−m in the equation 5x−y−4=0 We have, 5x−mx−5+m−4=0 ⇒‌‌(5−m)x+m−9=0 Therefore, x=‌
9−m
5−m
and y=m(‌
9−m
5−m
)+5−m ∴=‌
25−m
5−m
Here, Q=(‌
9−m
5−m
,‌
25−m
5−m
) Substituting y=mx+5−m in the equation 3x+4y−4=0 We have, 3x+4(mx+5−m)−4=0 (3+4m)x+16−4m=0 Therefore, x=‌
4m−16
4m+3
and y=‌
m(4m−16)
4m+3
+5−m y=‌
m+15
4m+3
and ‌y=‌
m(4m−16)
4m+3
+5−m ‌y=‌
m+15
4m+3
Hence, P=(‌
4m−16
4m+3
,‌
m+15
4m+3
) Since m(1 s) is the mid-point of PQ, we have 1‌=‌
1
2
[‌
9−m
5−m
+‌
4m−16
4m+3
]......(ii) and 5‌=‌
1
2
[‌
25−m
5−m
+‌
m+15
4m+3
].....(iii) From Eq. (ii), we get ‌2(5−m)(4m+3)