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Straight Lines and Pair of Straight Lines question

2022 · 27 Jul · Shift 2 · Q31
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Straight Lines and Pair of Straight Lines question

2022 · 27 Jul · Shift 2 · Q31

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
The equations of the sides AB,BC\mathrm{AB}, \mathrm{BC}AB,BC and CA of a triangle ABC are 2x+y=0,x+py=392 x+y=0, x+\mathrm{p} y=392x+y=0,x+py=39 and x−y=3x-y=3x−y=3 respectively and P(2,3)\mathrm{P}(2,3)P(2,3) is its circumcentre. Then which of the following is NOT true?
  1. A
    (AC)2=9p(\mathrm{AC})^{2}=9 \mathrm{p}(AC)2=9p
  2. B
    (AC)2+p2=136(\mathrm{AC})^{2}+\mathrm{p}^{2}=136(AC)2+p2=136
  3. C
    32<area⁡ (ΔABC)<3632\lt \operatorname{area}\,(\Delta \mathrm{ABC})\lt 3632<area(ΔABC)<36
  4. D
    34<area⁡ (△ABC)<3834\lt \operatorname{area}\,(\triangle \mathrm{ABC})\lt 3834<area(△ABC)<38
View written solutionFree

Correct answer: D

  1. Given sides of triangle

The sides are: AB:2x+y=0AB: 2x+y=0AB:2x+y=0 BC:x+py=39BC: x+py=39BC:x+py=39 CA:x−y=3CA: x-y=3CA:x−y=3

Circumcentre is P(2,3)P(2,3)P(2,3).

Since the circumcentre is equidistant from all three sides only in a special way? No — for a triangle, the circumcentre is the intersection of the perpendicular bisectors, so we should first find the vertices and then use the fact that PPP is equidistant from the vertices.

  1. Find vertices A,B,CA,B,CA,B,C

Vertex A=AB∩ACA = AB \cap ACA=AB∩AC

Solve

\quad x-y=3$$ From $x-y=3$, we get $y=x-3$. Substitute into $2x+y=0$: $$2x+(x-3)=0$$ $$3x=3$$ $$x=1,\, y=-2$$ So, $$A=(1,-2)$$ ### Vertex $B = AB \cap BC$ Solve $$2x+y=0, \quad x+py=39$$ From $2x+y=0$, we get $y=-2x$. Substitute into $x+py=39$: $$x+p(-2x)=39$$ $$x(1-2p)=39$$ $$x=\frac{39}{1-2p}, \quad y=-\frac{78}{1-2p}$$ So, $$B=\left(\frac{39}{1-2p},-\frac{78}{1-2p}\right)$$ ### Vertex $C = BC \cap AC$ Solve $$x+py=39, \quad x-y=3$$ From $x-y=3$, we get $x=y+3$. Substitute into $x+py=39$: $$(y+3)+py=39$$ $$(p+1)y=36$$ $$y=\frac{36}{p+1}, \quad x=\frac{36}{p+1}+3=\frac{3p+39}{p+1}$$ So, $$C=\left(\frac{3p+39}{p+1},\frac{36}{p+1}\right)$$ --- 3. **Use circumcentre condition** Since $P(2,3)$ is circumcentre, $$PA=PC$$ First compute $PA^2$: $$PA^2=(2-1)^2+(3-(-2))^2=1+25=26$$ Hence, $$PC^2=26$$ Now, $$C=\left(\frac{3p+39}{p+1},\frac{36}{p+1}\right)$$ So, $$\left(\frac{3p+39}{p+1}-2\right)^2+\left(\frac{36}{p+1}-3\right)^2=26$$ Simplify each term: $$\frac{3p+39}{p+1}-2=\frac{3p+39-2p-2}{p+1}=\frac{p+37}{p+1}$$ $$\frac{36}{p+1}-3=\frac{36-3p-3}{p+1}=\frac{33-3p}{p+1}=\frac{3(11-p)}{p+1}$$ Thus, $$\frac{(p+37)^2+9(11-p)^2}{(p+1)^2}=26$$ Expand numerator: $$(p+37)^2=p^2+74p+1369$$ $$9(11-p)^2=9(p^2-22p+121)=9p^2-198p+1089$$ So numerator is $$10p^2-124p+2458$$ Hence, $$10p^2-124p+2458=26(p+1)^2$$ $$10p^2-124p+2458=26p^2+52p+26$$ $$16p^2+176p-2432=0$$ $$p^2+11p-152=0$$ $$p=8 \quad \text{or} \quad p=-19$$ Now also use $PA=PB$. $$PB^2=\left(\frac{39}{1-2p}-2\right)^2+\left(-\frac{78}{1-2p}-3\right)^2$$ This must equal $26$. Check $p=8$: $$B=\left(\frac{39}{1-16},-\frac{78}{1-16}\right)=\left(-\frac{13}{5},\frac{26}{5}\right)$$ Then $$PB^2=\left(-\frac{13}{5}-2\right)^2+\left(\frac{26}{5}-3\right)^2 =\left(-\frac{23}{5}\right)^2+\left(\frac{11}{5}\right)^2 =\frac{529+121}{25}=26$$ Works. Check $p=-19$: $$B=\left(\frac{39}{39},-\frac{78}{39}\right)=(1,-2)=A$$ This makes the triangle degenerate, so reject. Therefore, $$\boxed{p=8}$$ --- 4. **Find $AC^2$** Since $A=(1,-2)$ and for $p=8$, $$C=\left(\frac{3\cdot 8+39}{8+1},\frac{36}{9}\right)=(7,4)$$ So, $$AC^2=(7-1)^2+(4-(-2))^2=6^2+6^2=72$$ Now check option A: $$9p=9\cdot 8=72$$ So, $$AC^2=9p$$ Hence **A is true**. --- 5. **Check option B** $$AC^2+p^2=72+8^2=72+64=136$$ Hence **B is true**. --- 6. **Find area of triangle** Vertices are: $$A=(1,-2),\quad C=(7,4)$$ And $$B=\left(-\frac{13}{5},\frac{26}{5}\right)$$ Use area formula: $$\text{Area}= \frac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$$ So, $$\text{Area}=\frac12\left|1\left(\frac{26}{5}-4\right)+\left(-\frac{13}{5}\right)(4-(-2))+7\left(-2-\frac{26}{5}\right)\right|$$ Compute terms: $$\frac{26}{5}-4=\frac{6}{5}$$ $$\left(-\frac{13}{5}\right)\cdot 6=-\frac{78}{5}$$ $$-2-\frac{26}{5}=-\frac{36}{5}, \quad 7\left(-\frac{36}{5}\right)=-\frac{252}{5}$$ Thus, $$\text{Area}=\frac12\left|\frac{6}{5}-\frac{78}{5}-\frac{252}{5}\right|$$ $$=\frac12\left|\frac{-324}{5}\right|$$ $$=\frac{162}{5}=32.4$$ --- 7. **Check options C and D** $$\text{Area}=32.4$$ - Option C: $32<\text{area}<36$ is **true**. - Option D: $34<\text{area}<38$ is **false**. Therefore, the statement which is **NOT true** is $$\boxed{D}$$ --- 8. **Comparison with stored correct answer** Stored correct answer: **D** Our derived answer: **D** So the answer agrees with the stored answer.
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