QBManager

Co-Ordinate system

88292 The centroid of the triangle formed by the lines \(x+y=1,2 x+3 y=6\) and \(4 x-y=-4\) lies in the quadrant

1 I

2 II

3 III

4 IV

CG PET-2016

Co-Ordinate system

88298 The distance of the origin from the centroid of the triangle whose two sides have the equations \(x-2 y+1=0\) and \(2 x-y-1=0\) and whose orthocenter is \(\left(\frac{7}{3}, \frac{7}{3}\right)\) is :

1 \(\sqrt{2}\)

2 2

3 \(2 \sqrt{2}\)

4 4

JEE Main-2022-29.06.2022

Co-Ordinate system

88293 Let the circumcentre of a triangle with vertices \(A(a, 3), B(b, 5)\) and \(C(a, b), a b>0\) be \(P(1,1)\). If the line AP intersects the line \(B C\) at the point \(Q\left(k_{1}, k_{2}\right)\), then \(k_{1}+k_{2}\) is equal to:

1 2

2 \(\frac{4}{7}\)

3 \(\frac{2}{7}\)

4 4

JEE Main-2022-29.07.2022

Co-Ordinate system

88294 Let \(A(1,1), B(-4,3), C(-2,-5)\) be vertices of a triangle \(A B C, P\) be a point on side \(B C\), and \(\Delta_{1}\) and \(\Delta_{2}\) be the areas of triangles \(A P B\) and \(A B C\), respectively. If \(\Delta_{1}: \Delta_{2}: 4: 7\), then the area enclosed by the lines \(A P, A C\) and the \(x\)-axis is

1 \(\frac{1}{4}\)

2 \(\frac{3}{4}\)

3 \(\frac{1}{2}\)

4 1

Explanation:

(C) : Given,
The co - ordinate of vestige of triangle are \(\mathrm{A}(1,1)\), B \((-4,3)\), C \((-2,-5)\)
We know that,

The Area of triangle \(\operatorname{ABC}\left(\Delta_{2}\right)=\frac{1}{2}\left|\begin{array}{ccc}1 1 1 \\ -4 3 1 \\ -2 -5 1\end{array}\right|\)
\(\Delta_{2}=\frac{1}{2}(8+2+26)=\frac{1}{2} \times 36\)
Similarly, The Area of triangle APB
\(\left(\Delta_{1}\right)=\frac{1}{2}\left|\begin{array}{ccc}x y 1 \\ 1 1 1 \\ -4 3 1\end{array}\right|\)
\(\Delta_{1}=\frac{1}{2}(\mathrm{x}(1-3)-\mathrm{y}(1+4)+1 \times 7)\)
\(\Delta_{1}=\frac{1}{2}(-2 \mathrm{x}-5 \mathrm{y}+7)\)
\(\therefore \frac{\Delta_{1}}{\Delta_{2}}=\frac{(-2 \mathrm{x}-5 \mathrm{y}+7)}{36}=\frac{4}{7}\) \((-2 x-5 y+7)=36 \times 4\)
\(-14 x-35 y+49=144\)
\(-14 x-25 y=95\)
\(14 \mathrm{x}+35 \mathrm{y}=-95\)
\(\therefore\) Equation of line \(\mathrm{BC}\) is
\((y-3)=\frac{-5-3}{-2+4}(x+4)\)
\(\mathrm{y}-3=\frac{-8}{2}(\mathrm{x}+4)\)
\(y-3=-4(x+4)\)
\(y-3=-4 x-16\)
\(4 \mathrm{x}+\mathrm{y}+13=0\)
\(\therefore\) From equation (i) and (ii), we get-
\(14 x+35 y =-95 \tag{ii}\)
\(140 x+35 y =-13 \times 35\)
\(-126 x =5(13 \times 7-19)\)
\(=5(91-19)\)
\(126 x=72 \times\)
\(x=-\frac{72 \times 5}{126}\)
\(x=\frac{-20}{7}\)
\(\therefore y=-13+\frac{4 \times 20}{7}, \quad y=\frac{-11}{7}\)
\(\therefore\) Co - ordinate of point \(\mathrm{P}\left(\frac{-20}{7}, \frac{-11}{7}\right)\)
Now, equation of line AP is
\((y-1)=\frac{1+11 / 7}{1+20 / 7}(x-1)\)
\(y-1=\frac{18}{27}(x-1)\)
Put \(\mathrm{y}=0\)
\(-27=18(\mathrm{x}-1)\)
\(-3=2 x-2\)
\(\mathrm{x}=-1 / 2\)
\(\therefore\) Co - coordinate of point \(\mathrm{R}\left(-\frac{1}{2}, 0\right)\)
Similarly, The equation of line AB
\(\mathrm{y}-1=\frac{1+5}{1+2}(\mathrm{x}-1)\)
\(\mathrm{y}-1=\frac{6}{3}(\mathrm{x}-1)\)
\(\therefore\) Put \(\mathrm{y}=0\)
\(-1=2 \mathrm{x}-2\)
\(\mathrm{x}=\frac{1}{2}\)
\(\therefore\) The co - ordinate of point \(\mathrm{s}\left(\frac{1}{2}, 0\right)\)
Now, The area of triangle ARS \(=\frac{1}{2} \times 1 \times 1=\frac{1}{2}\)

JEE Main-27.07.2022

Co-Ordinate system

88295 Consider two points \(A(a, b)\) and \(B(c, d)\) where \(c>a>0\) and \(d>b>0\), then abscissa of a point on \(x\)-axis such that sum of its distances from \(A\) and \(B\) is minimum, is :

1 \((b c-a d) /(b-d)\)

2 \((a c+b d) /(b+d)\)

3 \((\mathrm{ac}-\mathrm{bd}) /(\mathrm{b}-\mathrm{a})\)

4 \((a d+b c) /(b+d)\)

SCRA-2010

Co-Ordinate system

88292 The centroid of the triangle formed by the lines \(x+y=1,2 x+3 y=6\) and \(4 x-y=-4\) lies in the quadrant

1 I

2 II

3 III

4 IV

CG PET-2016

Co-Ordinate system

88298 The distance of the origin from the centroid of the triangle whose two sides have the equations \(x-2 y+1=0\) and \(2 x-y-1=0\) and whose orthocenter is \(\left(\frac{7}{3}, \frac{7}{3}\right)\) is :

1 \(\sqrt{2}\)

2 2

3 \(2 \sqrt{2}\)

4 4

JEE Main-2022-29.06.2022

Co-Ordinate system

88293 Let the circumcentre of a triangle with vertices \(A(a, 3), B(b, 5)\) and \(C(a, b), a b>0\) be \(P(1,1)\). If the line AP intersects the line \(B C\) at the point \(Q\left(k_{1}, k_{2}\right)\), then \(k_{1}+k_{2}\) is equal to:

1 2

2 \(\frac{4}{7}\)

3 \(\frac{2}{7}\)

4 4

JEE Main-2022-29.07.2022

Co-Ordinate system

88294 Let \(A(1,1), B(-4,3), C(-2,-5)\) be vertices of a triangle \(A B C, P\) be a point on side \(B C\), and \(\Delta_{1}\) and \(\Delta_{2}\) be the areas of triangles \(A P B\) and \(A B C\), respectively. If \(\Delta_{1}: \Delta_{2}: 4: 7\), then the area enclosed by the lines \(A P, A C\) and the \(x\)-axis is

1 \(\frac{1}{4}\)

2 \(\frac{3}{4}\)

3 \(\frac{1}{2}\)

4 1

Explanation:

(C) : Given,
The co - ordinate of vestige of triangle are \(\mathrm{A}(1,1)\), B \((-4,3)\), C \((-2,-5)\)
We know that,

The Area of triangle \(\operatorname{ABC}\left(\Delta_{2}\right)=\frac{1}{2}\left|\begin{array}{ccc}1 1 1 \\ -4 3 1 \\ -2 -5 1\end{array}\right|\)
\(\Delta_{2}=\frac{1}{2}(8+2+26)=\frac{1}{2} \times 36\)
Similarly, The Area of triangle APB
\(\left(\Delta_{1}\right)=\frac{1}{2}\left|\begin{array}{ccc}x y 1 \\ 1 1 1 \\ -4 3 1\end{array}\right|\)
\(\Delta_{1}=\frac{1}{2}(\mathrm{x}(1-3)-\mathrm{y}(1+4)+1 \times 7)\)
\(\Delta_{1}=\frac{1}{2}(-2 \mathrm{x}-5 \mathrm{y}+7)\)
\(\therefore \frac{\Delta_{1}}{\Delta_{2}}=\frac{(-2 \mathrm{x}-5 \mathrm{y}+7)}{36}=\frac{4}{7}\) \((-2 x-5 y+7)=36 \times 4\)
\(-14 x-35 y+49=144\)
\(-14 x-25 y=95\)
\(14 \mathrm{x}+35 \mathrm{y}=-95\)
\(\therefore\) Equation of line \(\mathrm{BC}\) is
\((y-3)=\frac{-5-3}{-2+4}(x+4)\)
\(\mathrm{y}-3=\frac{-8}{2}(\mathrm{x}+4)\)
\(y-3=-4(x+4)\)
\(y-3=-4 x-16\)
\(4 \mathrm{x}+\mathrm{y}+13=0\)
\(\therefore\) From equation (i) and (ii), we get-
\(14 x+35 y =-95 \tag{ii}\)
\(140 x+35 y =-13 \times 35\)
\(-126 x =5(13 \times 7-19)\)
\(=5(91-19)\)
\(126 x=72 \times\)
\(x=-\frac{72 \times 5}{126}\)
\(x=\frac{-20}{7}\)
\(\therefore y=-13+\frac{4 \times 20}{7}, \quad y=\frac{-11}{7}\)
\(\therefore\) Co - ordinate of point \(\mathrm{P}\left(\frac{-20}{7}, \frac{-11}{7}\right)\)
Now, equation of line AP is
\((y-1)=\frac{1+11 / 7}{1+20 / 7}(x-1)\)
\(y-1=\frac{18}{27}(x-1)\)
Put \(\mathrm{y}=0\)
\(-27=18(\mathrm{x}-1)\)
\(-3=2 x-2\)
\(\mathrm{x}=-1 / 2\)
\(\therefore\) Co - coordinate of point \(\mathrm{R}\left(-\frac{1}{2}, 0\right)\)
Similarly, The equation of line AB
\(\mathrm{y}-1=\frac{1+5}{1+2}(\mathrm{x}-1)\)
\(\mathrm{y}-1=\frac{6}{3}(\mathrm{x}-1)\)
\(\therefore\) Put \(\mathrm{y}=0\)
\(-1=2 \mathrm{x}-2\)
\(\mathrm{x}=\frac{1}{2}\)
\(\therefore\) The co - ordinate of point \(\mathrm{s}\left(\frac{1}{2}, 0\right)\)
Now, The area of triangle ARS \(=\frac{1}{2} \times 1 \times 1=\frac{1}{2}\)

JEE Main-27.07.2022

Co-Ordinate system

88295 Consider two points \(A(a, b)\) and \(B(c, d)\) where \(c>a>0\) and \(d>b>0\), then abscissa of a point on \(x\)-axis such that sum of its distances from \(A\) and \(B\) is minimum, is :

1 \((b c-a d) /(b-d)\)

2 \((a c+b d) /(b+d)\)

3 \((\mathrm{ac}-\mathrm{bd}) /(\mathrm{b}-\mathrm{a})\)

4 \((a d+b c) /(b+d)\)

SCRA-2010

Co-Ordinate system

88292 The centroid of the triangle formed by the lines \(x+y=1,2 x+3 y=6\) and \(4 x-y=-4\) lies in the quadrant

1 I

2 II

3 III

4 IV

CG PET-2016

Co-Ordinate system

88298 The distance of the origin from the centroid of the triangle whose two sides have the equations \(x-2 y+1=0\) and \(2 x-y-1=0\) and whose orthocenter is \(\left(\frac{7}{3}, \frac{7}{3}\right)\) is :

1 \(\sqrt{2}\)

2 2

3 \(2 \sqrt{2}\)

4 4

JEE Main-2022-29.06.2022

Co-Ordinate system

88293 Let the circumcentre of a triangle with vertices \(A(a, 3), B(b, 5)\) and \(C(a, b), a b>0\) be \(P(1,1)\). If the line AP intersects the line \(B C\) at the point \(Q\left(k_{1}, k_{2}\right)\), then \(k_{1}+k_{2}\) is equal to:

1 2

2 \(\frac{4}{7}\)

3 \(\frac{2}{7}\)

4 4

JEE Main-2022-29.07.2022

Co-Ordinate system

88294 Let \(A(1,1), B(-4,3), C(-2,-5)\) be vertices of a triangle \(A B C, P\) be a point on side \(B C\), and \(\Delta_{1}\) and \(\Delta_{2}\) be the areas of triangles \(A P B\) and \(A B C\), respectively. If \(\Delta_{1}: \Delta_{2}: 4: 7\), then the area enclosed by the lines \(A P, A C\) and the \(x\)-axis is

1 \(\frac{1}{4}\)

2 \(\frac{3}{4}\)

3 \(\frac{1}{2}\)

4 1

Explanation:

(C) : Given,
The co - ordinate of vestige of triangle are \(\mathrm{A}(1,1)\), B \((-4,3)\), C \((-2,-5)\)
We know that,

The Area of triangle \(\operatorname{ABC}\left(\Delta_{2}\right)=\frac{1}{2}\left|\begin{array}{ccc}1 1 1 \\ -4 3 1 \\ -2 -5 1\end{array}\right|\)
\(\Delta_{2}=\frac{1}{2}(8+2+26)=\frac{1}{2} \times 36\)
Similarly, The Area of triangle APB
\(\left(\Delta_{1}\right)=\frac{1}{2}\left|\begin{array}{ccc}x y 1 \\ 1 1 1 \\ -4 3 1\end{array}\right|\)
\(\Delta_{1}=\frac{1}{2}(\mathrm{x}(1-3)-\mathrm{y}(1+4)+1 \times 7)\)
\(\Delta_{1}=\frac{1}{2}(-2 \mathrm{x}-5 \mathrm{y}+7)\)
\(\therefore \frac{\Delta_{1}}{\Delta_{2}}=\frac{(-2 \mathrm{x}-5 \mathrm{y}+7)}{36}=\frac{4}{7}\) \((-2 x-5 y+7)=36 \times 4\)
\(-14 x-35 y+49=144\)
\(-14 x-25 y=95\)
\(14 \mathrm{x}+35 \mathrm{y}=-95\)
\(\therefore\) Equation of line \(\mathrm{BC}\) is
\((y-3)=\frac{-5-3}{-2+4}(x+4)\)
\(\mathrm{y}-3=\frac{-8}{2}(\mathrm{x}+4)\)
\(y-3=-4(x+4)\)
\(y-3=-4 x-16\)
\(4 \mathrm{x}+\mathrm{y}+13=0\)
\(\therefore\) From equation (i) and (ii), we get-
\(14 x+35 y =-95 \tag{ii}\)
\(140 x+35 y =-13 \times 35\)
\(-126 x =5(13 \times 7-19)\)
\(=5(91-19)\)
\(126 x=72 \times\)
\(x=-\frac{72 \times 5}{126}\)
\(x=\frac{-20}{7}\)
\(\therefore y=-13+\frac{4 \times 20}{7}, \quad y=\frac{-11}{7}\)
\(\therefore\) Co - ordinate of point \(\mathrm{P}\left(\frac{-20}{7}, \frac{-11}{7}\right)\)
Now, equation of line AP is
\((y-1)=\frac{1+11 / 7}{1+20 / 7}(x-1)\)
\(y-1=\frac{18}{27}(x-1)\)
Put \(\mathrm{y}=0\)
\(-27=18(\mathrm{x}-1)\)
\(-3=2 x-2\)
\(\mathrm{x}=-1 / 2\)
\(\therefore\) Co - coordinate of point \(\mathrm{R}\left(-\frac{1}{2}, 0\right)\)
Similarly, The equation of line AB
\(\mathrm{y}-1=\frac{1+5}{1+2}(\mathrm{x}-1)\)
\(\mathrm{y}-1=\frac{6}{3}(\mathrm{x}-1)\)
\(\therefore\) Put \(\mathrm{y}=0\)
\(-1=2 \mathrm{x}-2\)
\(\mathrm{x}=\frac{1}{2}\)
\(\therefore\) The co - ordinate of point \(\mathrm{s}\left(\frac{1}{2}, 0\right)\)
Now, The area of triangle ARS \(=\frac{1}{2} \times 1 \times 1=\frac{1}{2}\)

JEE Main-27.07.2022

Co-Ordinate system

88295 Consider two points \(A(a, b)\) and \(B(c, d)\) where \(c>a>0\) and \(d>b>0\), then abscissa of a point on \(x\)-axis such that sum of its distances from \(A\) and \(B\) is minimum, is :

1 \((b c-a d) /(b-d)\)

2 \((a c+b d) /(b+d)\)

3 \((\mathrm{ac}-\mathrm{bd}) /(\mathrm{b}-\mathrm{a})\)

4 \((a d+b c) /(b+d)\)

SCRA-2010

Co-Ordinate system

88292 The centroid of the triangle formed by the lines \(x+y=1,2 x+3 y=6\) and \(4 x-y=-4\) lies in the quadrant

1 I

2 II

3 III

4 IV

CG PET-2016

Co-Ordinate system

88298 The distance of the origin from the centroid of the triangle whose two sides have the equations \(x-2 y+1=0\) and \(2 x-y-1=0\) and whose orthocenter is \(\left(\frac{7}{3}, \frac{7}{3}\right)\) is :

1 \(\sqrt{2}\)

2 2

3 \(2 \sqrt{2}\)

4 4

JEE Main-2022-29.06.2022

Co-Ordinate system

88293 Let the circumcentre of a triangle with vertices \(A(a, 3), B(b, 5)\) and \(C(a, b), a b>0\) be \(P(1,1)\). If the line AP intersects the line \(B C\) at the point \(Q\left(k_{1}, k_{2}\right)\), then \(k_{1}+k_{2}\) is equal to:

1 2

2 \(\frac{4}{7}\)

3 \(\frac{2}{7}\)

4 4

JEE Main-2022-29.07.2022

Co-Ordinate system

88294 Let \(A(1,1), B(-4,3), C(-2,-5)\) be vertices of a triangle \(A B C, P\) be a point on side \(B C\), and \(\Delta_{1}\) and \(\Delta_{2}\) be the areas of triangles \(A P B\) and \(A B C\), respectively. If \(\Delta_{1}: \Delta_{2}: 4: 7\), then the area enclosed by the lines \(A P, A C\) and the \(x\)-axis is

1 \(\frac{1}{4}\)

2 \(\frac{3}{4}\)

3 \(\frac{1}{2}\)

4 1

Explanation:

(C) : Given,
The co - ordinate of vestige of triangle are \(\mathrm{A}(1,1)\), B \((-4,3)\), C \((-2,-5)\)
We know that,

The Area of triangle \(\operatorname{ABC}\left(\Delta_{2}\right)=\frac{1}{2}\left|\begin{array}{ccc}1 1 1 \\ -4 3 1 \\ -2 -5 1\end{array}\right|\)
\(\Delta_{2}=\frac{1}{2}(8+2+26)=\frac{1}{2} \times 36\)
Similarly, The Area of triangle APB
\(\left(\Delta_{1}\right)=\frac{1}{2}\left|\begin{array}{ccc}x y 1 \\ 1 1 1 \\ -4 3 1\end{array}\right|\)
\(\Delta_{1}=\frac{1}{2}(\mathrm{x}(1-3)-\mathrm{y}(1+4)+1 \times 7)\)
\(\Delta_{1}=\frac{1}{2}(-2 \mathrm{x}-5 \mathrm{y}+7)\)
\(\therefore \frac{\Delta_{1}}{\Delta_{2}}=\frac{(-2 \mathrm{x}-5 \mathrm{y}+7)}{36}=\frac{4}{7}\) \((-2 x-5 y+7)=36 \times 4\)
\(-14 x-35 y+49=144\)
\(-14 x-25 y=95\)
\(14 \mathrm{x}+35 \mathrm{y}=-95\)
\(\therefore\) Equation of line \(\mathrm{BC}\) is
\((y-3)=\frac{-5-3}{-2+4}(x+4)\)
\(\mathrm{y}-3=\frac{-8}{2}(\mathrm{x}+4)\)
\(y-3=-4(x+4)\)
\(y-3=-4 x-16\)
\(4 \mathrm{x}+\mathrm{y}+13=0\)
\(\therefore\) From equation (i) and (ii), we get-
\(14 x+35 y =-95 \tag{ii}\)
\(140 x+35 y =-13 \times 35\)
\(-126 x =5(13 \times 7-19)\)
\(=5(91-19)\)
\(126 x=72 \times\)
\(x=-\frac{72 \times 5}{126}\)
\(x=\frac{-20}{7}\)
\(\therefore y=-13+\frac{4 \times 20}{7}, \quad y=\frac{-11}{7}\)
\(\therefore\) Co - ordinate of point \(\mathrm{P}\left(\frac{-20}{7}, \frac{-11}{7}\right)\)
Now, equation of line AP is
\((y-1)=\frac{1+11 / 7}{1+20 / 7}(x-1)\)
\(y-1=\frac{18}{27}(x-1)\)
Put \(\mathrm{y}=0\)
\(-27=18(\mathrm{x}-1)\)
\(-3=2 x-2\)
\(\mathrm{x}=-1 / 2\)
\(\therefore\) Co - coordinate of point \(\mathrm{R}\left(-\frac{1}{2}, 0\right)\)
Similarly, The equation of line AB
\(\mathrm{y}-1=\frac{1+5}{1+2}(\mathrm{x}-1)\)
\(\mathrm{y}-1=\frac{6}{3}(\mathrm{x}-1)\)
\(\therefore\) Put \(\mathrm{y}=0\)
\(-1=2 \mathrm{x}-2\)
\(\mathrm{x}=\frac{1}{2}\)
\(\therefore\) The co - ordinate of point \(\mathrm{s}\left(\frac{1}{2}, 0\right)\)
Now, The area of triangle ARS \(=\frac{1}{2} \times 1 \times 1=\frac{1}{2}\)

JEE Main-27.07.2022

Co-Ordinate system

88295 Consider two points \(A(a, b)\) and \(B(c, d)\) where \(c>a>0\) and \(d>b>0\), then abscissa of a point on \(x\)-axis such that sum of its distances from \(A\) and \(B\) is minimum, is :

1 \((b c-a d) /(b-d)\)

2 \((a c+b d) /(b+d)\)

3 \((\mathrm{ac}-\mathrm{bd}) /(\mathrm{b}-\mathrm{a})\)

4 \((a d+b c) /(b+d)\)

SCRA-2010

Co-Ordinate system

88292 The centroid of the triangle formed by the lines \(x+y=1,2 x+3 y=6\) and \(4 x-y=-4\) lies in the quadrant

1 I

2 II

3 III

4 IV

CG PET-2016

Co-Ordinate system

88298 The distance of the origin from the centroid of the triangle whose two sides have the equations \(x-2 y+1=0\) and \(2 x-y-1=0\) and whose orthocenter is \(\left(\frac{7}{3}, \frac{7}{3}\right)\) is :

1 \(\sqrt{2}\)

2 2

3 \(2 \sqrt{2}\)

4 4

JEE Main-2022-29.06.2022

Co-Ordinate system

88293 Let the circumcentre of a triangle with vertices \(A(a, 3), B(b, 5)\) and \(C(a, b), a b>0\) be \(P(1,1)\). If the line AP intersects the line \(B C\) at the point \(Q\left(k_{1}, k_{2}\right)\), then \(k_{1}+k_{2}\) is equal to:

1 2

2 \(\frac{4}{7}\)

3 \(\frac{2}{7}\)

4 4

JEE Main-2022-29.07.2022

Co-Ordinate system

88294 Let \(A(1,1), B(-4,3), C(-2,-5)\) be vertices of a triangle \(A B C, P\) be a point on side \(B C\), and \(\Delta_{1}\) and \(\Delta_{2}\) be the areas of triangles \(A P B\) and \(A B C\), respectively. If \(\Delta_{1}: \Delta_{2}: 4: 7\), then the area enclosed by the lines \(A P, A C\) and the \(x\)-axis is

1 \(\frac{1}{4}\)

2 \(\frac{3}{4}\)

3 \(\frac{1}{2}\)

4 1

Explanation:

(C) : Given,
The co - ordinate of vestige of triangle are \(\mathrm{A}(1,1)\), B \((-4,3)\), C \((-2,-5)\)
We know that,

The Area of triangle \(\operatorname{ABC}\left(\Delta_{2}\right)=\frac{1}{2}\left|\begin{array}{ccc}1 1 1 \\ -4 3 1 \\ -2 -5 1\end{array}\right|\)
\(\Delta_{2}=\frac{1}{2}(8+2+26)=\frac{1}{2} \times 36\)
Similarly, The Area of triangle APB
\(\left(\Delta_{1}\right)=\frac{1}{2}\left|\begin{array}{ccc}x y 1 \\ 1 1 1 \\ -4 3 1\end{array}\right|\)
\(\Delta_{1}=\frac{1}{2}(\mathrm{x}(1-3)-\mathrm{y}(1+4)+1 \times 7)\)
\(\Delta_{1}=\frac{1}{2}(-2 \mathrm{x}-5 \mathrm{y}+7)\)
\(\therefore \frac{\Delta_{1}}{\Delta_{2}}=\frac{(-2 \mathrm{x}-5 \mathrm{y}+7)}{36}=\frac{4}{7}\) \((-2 x-5 y+7)=36 \times 4\)
\(-14 x-35 y+49=144\)
\(-14 x-25 y=95\)
\(14 \mathrm{x}+35 \mathrm{y}=-95\)
\(\therefore\) Equation of line \(\mathrm{BC}\) is
\((y-3)=\frac{-5-3}{-2+4}(x+4)\)
\(\mathrm{y}-3=\frac{-8}{2}(\mathrm{x}+4)\)
\(y-3=-4(x+4)\)
\(y-3=-4 x-16\)
\(4 \mathrm{x}+\mathrm{y}+13=0\)
\(\therefore\) From equation (i) and (ii), we get-
\(14 x+35 y =-95 \tag{ii}\)
\(140 x+35 y =-13 \times 35\)
\(-126 x =5(13 \times 7-19)\)
\(=5(91-19)\)
\(126 x=72 \times\)
\(x=-\frac{72 \times 5}{126}\)
\(x=\frac{-20}{7}\)
\(\therefore y=-13+\frac{4 \times 20}{7}, \quad y=\frac{-11}{7}\)
\(\therefore\) Co - ordinate of point \(\mathrm{P}\left(\frac{-20}{7}, \frac{-11}{7}\right)\)
Now, equation of line AP is
\((y-1)=\frac{1+11 / 7}{1+20 / 7}(x-1)\)
\(y-1=\frac{18}{27}(x-1)\)
Put \(\mathrm{y}=0\)
\(-27=18(\mathrm{x}-1)\)
\(-3=2 x-2\)
\(\mathrm{x}=-1 / 2\)
\(\therefore\) Co - coordinate of point \(\mathrm{R}\left(-\frac{1}{2}, 0\right)\)
Similarly, The equation of line AB
\(\mathrm{y}-1=\frac{1+5}{1+2}(\mathrm{x}-1)\)
\(\mathrm{y}-1=\frac{6}{3}(\mathrm{x}-1)\)
\(\therefore\) Put \(\mathrm{y}=0\)
\(-1=2 \mathrm{x}-2\)
\(\mathrm{x}=\frac{1}{2}\)
\(\therefore\) The co - ordinate of point \(\mathrm{s}\left(\frac{1}{2}, 0\right)\)
Now, The area of triangle ARS \(=\frac{1}{2} \times 1 \times 1=\frac{1}{2}\)

JEE Main-27.07.2022

Co-Ordinate system

88295 Consider two points \(A(a, b)\) and \(B(c, d)\) where \(c>a>0\) and \(d>b>0\), then abscissa of a point on \(x\)-axis such that sum of its distances from \(A\) and \(B\) is minimum, is :

1 \((b c-a d) /(b-d)\)

2 \((a c+b d) /(b+d)\)

3 \((\mathrm{ac}-\mathrm{bd}) /(\mathrm{b}-\mathrm{a})\)

4 \((a d+b c) /(b+d)\)

SCRA-2010