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Straight Lines and Pair of Straight Lines question

2023 · 24 Jan · Shift 2 · Q41
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Straight Lines and Pair of Straight Lines question

2023 · 24 Jan · Shift 2 · Q41

JEE MainMathematicsStraight Lines and Pair of Straight LinesNumerical+4 / −1
The equations of the sides AB, BC and CA of a triangle ABC are : 2x+y=0,x+py=21a,(a±0)2x+y=0,x+py=21a,(a\pm0)2x+y=0,x+py=21a,(a±0) and x−y=3x-y=3x−y=3 respectively. Let P(2, a) be the centroid of Δ\DeltaΔ ABC. Then (BC) 2^22 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 122

We are given the three sides of triangle ABCABCABC:

  • AB:2x+y=0AB: 2x+y=0AB:2x+y=0
  • BC:x+py=21aBC: x+py=21aBC:x+py=21a
  • CA:x−y=3CA: x-y=3CA:x−y=3

and the centroid is P(2,a)P(2,a)P(2,a).

We need to find (BC)2(BC)^2(BC)2.

1. Find the vertices from intersection of sides

Vertex A=AB∩ACA = AB \cap ACA=AB∩AC

Solve 2x+y=0,2x+y=0,2x+y=0, x−y=3.x-y=3.x−y=3.

From x−y=3x-y=3x−y=3, we get y=x−3.y=x-3.y=x−3. Substitute into 2x+y=02x+y=02x+y=0: 2x+(x−3)=02x+(x-3)=02x+(x−3)=0 3x−3=03x-3=03x−3=0 x=1, y=−2.x=1,\, y=-2.x=1,y=−2.

So, A=(1,−2).A=(1,-2).A=(1,−2).

Vertex B=AB∩BCB = AB \cap BCB=AB∩BC

Solve 2x+y=0⇒y=−2x,2x+y=0 \Rightarrow y=-2x,2x+y=0⇒y=−2x, x+py=21a.x+py=21a.x+py=21a.

Substitute y=−2xy=-2xy=−2x into the second equation: x+p(−2x)=21ax+p(-2x)=21ax+p(−2x)=21a (1−2p)x=21a.(1-2p)x=21a.(1−2p)x=21a.

Hence xB=21a1−2p,yB=−42a1−2p.x_B=\frac{21a}{1-2p}, \qquad y_B=-\frac{42a}{1-2p}.xB​=1−2p21a​,yB​=−1−2p42a​.

So, B=(21a1−2p,−42a1−2p).B=\left(\frac{21a}{1-2p},-\frac{42a}{1-2p}\right).B=(1−2p21a​,−1−2p42a​).

Vertex C=BC∩CAC = BC \cap CAC=BC∩CA

Solve x−y=3⇒x=y+3,x-y=3 \Rightarrow x=y+3,x−y=3⇒x=y+3, x+py=21a.x+py=21a.x+py=21a.

Substitute x=y+3x=y+3x=y+3: (y+3)+py=21a(y+3)+py=21a(y+3)+py=21a (1+p)y=21a−3(1+p)y=21a-3(1+p)y=21a−3 yC=21a−31+p.y_C=\frac{21a-3}{1+p}.yC​=1+p21a−3​.

Then

=\frac{21a-3+3+3p}{1+p} =\frac{21a+3p}{1+p}.$$ So, $$C=\left(\frac{21a+3p}{1+p},\frac{21a-3}{1+p}\right).$$ --- ## 2. Use centroid formula Centroid of triangle is $$\left(\frac{x_A+x_B+x_C}{3},\frac{y_A+y_B+y_C}{3}\right)=(2,a).$$ Since $A=(1,-2)$, we get two equations. ### From x-coordinate: $$\frac{1+\frac{21a}{1-2p}+\frac{21a+3p}{1+p}}{3}=2.$$ So, $$1+\frac{21a}{1-2p}+\frac{21a+3p}{1+p}=6,$$ $$\frac{21a}{1-2p}+\frac{21a+3p}{1+p}=5. \quad (1)$$ ### From y-coordinate: $$\frac{-2-\frac{42a}{1-2p}+\frac{21a-3}{1+p}}{3}=a.$$ So, $$-2-\frac{42a}{1-2p}+\frac{21a-3}{1+p}=3a. \quad (2)$$ --- ## 3. Solve for $a$ and $p$ Let $$u=\frac{1}{1-2p}, \qquad v=\frac{1}{1+p}.$$ Then equations become: From (1): $$21au+(21a+3p)v=5.$$ From (2): $$-2-42au+(21a-3)v=3a.$$ Instead of continuing in $u,v$, we solve directly by trying to simplify using the centroid condition in a cleaner way. --- ## 4. Write $B$ and $C$ using section parameters From line $AB: 2x+y=0$, every point on it is of form $(t,-2t)$. Thus let $$B=(m,-2m).$$ Since $B$ lies on $BC: x+py=21a$, $$m+p(-2m)=21a$$ $$m(1-2p)=21a. \quad (3)$$ From line $CA: x-y=3$, every point on it is of form $(n,n-3)$. Thus let $$C=(n,n-3).$$ Since $C$ lies on $BC: x+py=21a$, $$n+p(n-3)=21a$$ $$n(1+p)-3p=21a. \quad (4)$$ Now use centroid: $$\left(\frac{1+m+n}{3},\frac{-2-2m+(n-3)}{3}\right)=(2,a).$$ So, $$1+m+n=6 \Rightarrow m+n=5. \quad (5)$$ And $$-2-2m+n-3=3a$$ $$n-2m-5=3a. \quad (6)$$ From (5), $$n=5-m.$$ Substitute into (6): $$(5-m)-2m-5=3a$$ $$-3m=3a$$ $$m=-a.$$ Then from $m+n=5$, $$n=5+a.$$ So, $$B=(-a,2a), \qquad C=(5+a, a+2).$$ --- ## 5. Use the fact that both $B$ and $C$ lie on line $BC: x+py=21a$ For point $B=(-a,2a)$: $$-a+2ap=21a.$$ Since $a\ne 0$, $$-1+2p=21$$ $$2p=22$$ $$p=11.$$ For point $C=(5+a,a+2)$: $$(5+a)+11(a+2)=21a$$ $$5+a+11a+22=21a$$ $$27+12a=21a$$ $$9a=27$$ $$a=3.$$ Thus, $$B=(-3,6), \qquad C=(8,5).$$ --- ## 6. Compute $(BC)^2$ $$BC^2=(8-(-3))^2+(5-6)^2$$ $$=11^2+(-1)^2$$ $$=121+1$$ $$=122.$$ --- ## 7. Final answer $$\boxed{122}$$ The derived answer matches the stored correct answer.
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