12. If A (1,-2), B (-2.3) and C (2.-5) are the vertices of AABC, then

1.	12. If A (1,-2), B (-2.3) and C (2.-5) are the vertices of AABC, then the equation of the median BE is(a) 7x + 13y + 47 = 0 (b) 13x + 7y + 5 = 0(c) 7x – 13y + 5 = 0 (d) none of these
Answer» GIVEN :- The vertices of triangle ABC as A (1, -2) B (-2, 3) C (2, -5) To Find :- Equation of median BE Solution :- We KNOW, Median BE divides the opposite side AC in to two equal parts. It implies, E is the midpoint of AC. Coordinates of A = (1, - 2) Coordinates of C = (2, - 5) Using midpoint Formula, ${\underline{\boxed{\rm \: (x,y) = \bigg({ \dfrac{x_1 + x_2}{2} \; ,\; \dfrac{y_1 + y_2}{2} \bigg) \quad}}}}$ Here, • x₁ = 1 • x₂ = 2 • y₁ = - 2 • y₂ = - 5 Hence, Coordinates of E is evaluated as $\rm :\longmapsto\:{\rm \: Coordinates \: of \: E = \bigg({ \dfrac{1 +2}{2} \; ,\; \dfrac{ - 2 - 5}{2} \bigg) \quad}}$ ${\underline{\boxed{\rm \: Coordinates \: of \: E = \bigg({\dfrac{3}{2} \; ,\; \dfrac{ -7}{2} \bigg) \quad}}}}$ Now, We know, Equation of line passing through 2 points is given by $\rm :\longmapsto\:\bf \:y - y_1 = \dfrac{y_2-y_1}{x_2-x_1} (x-x_1)$ Now, $\rm :\longmapsto\:{\rm \: Coordinates \: of \: E = \bigg({\dfrac{3}{2} \; ,\; \dfrac{ -7}{2} \bigg) \quad}}$ $\rm :\longmapsto\:Coordinates \: of \: B = ( - 2, \: 3)$ Here, $\rm :\longmapsto\:x_1 = - 2,y_1 = 3,x_2= \dfrac{3}{2} ,y_2= \dfrac{ - 7}{2}$ Hence, Equation of line BE is $\rm :\longmapsto\:\:y - 3 = \dfrac{\dfrac{ - 7}{2} - 3}{\dfrac{3}{2} + 2 } (x + 2)$ $\rm :\longmapsto\:\:y - 3 = \dfrac{\dfrac{ - 7 - 6}{2}}{\dfrac{3 + 4}{2}} (x + 2)$ $\rm :\longmapsto\:\:y - 3 = \dfrac{ - 13}{7} (x + 2)$ $\rm :\longmapsto\:7y - 21 = - 13x - 26$ $\bf\implies \:13x + 7y + 5 = 0$ $\large{\boxed{\boxed{\bf{Option \: (b) \: is \: correct}}}}$ Additional Information :- Different forms of equations of a STRAIGHT line 1. Equations of horizontal and vertical lines Equation of the lines which are horizontal or parallel to the X-axis is y = a, where a is the y – coordinate of the points on the line. Similarly, equation of a straight line which is vertical or parallel to Y-axis is x = a, where a is the x-coordinate of the points on the line 2. Point-slope form equation of line Consider a non-vertical line L whose slope is m, A(x,y) be an arbitrary point on the line and P(a, b) be the fixed point on the same line. Equation of line is given by y - b = m(x - a) 3. Slope-intercept form equation of line Consider a line whose slope is m which CUTS the Y-axis at a distance ‘a’ from the ORIGIN. Then the distance a is called the y– intercept of the line. The point at which the line cuts y-axis will be (0,a). Then equation of line is given by y = mx + a. 4. Intercept Form of Line Consider a line L having x– intercept a and y– intercept b, then the line passes through X– axis at (a,0) and Y– axis at (0,b). Equation of line is given by x/a + y/b = 1. 5. Normal form of Line Consider a perpendicular from the origin having length p to line L and it makes an angle β with the positive X-axis. Then, the equation of line is given by x cosβ + y sinβ = p.

12. If A (1,-2), B (-2.3) and C (2.-5) are the vertices of AABC, then the equation of the median BE is(a) 7x + 13y + 47 = 0 (b) 13x + 7y + 5 = 0(c) 7x – 13y + 5 = 0 (d) none of these

Answer»

GIVEN :-

The vertices of triangle ABC as

A (1, -2)

B (-2, 3)

C (2, -5)

To Find :-

Equation of median BE

Solution :-

We KNOW,

Median BE divides the opposite side AC in to two equal parts.

It implies, E is the midpoint of AC.

Coordinates of A = (1, - 2)

Coordinates of C = (2, - 5)

Using midpoint Formula,

${\underline{\boxed{\rm \: (x,y) = \bigg({ \dfrac{x_1 + x_2}{2} \; ,\; \dfrac{y_1 + y_2}{2} \bigg) \quad}}}}$

Here,

• x₁ = 1

• x₂ = 2

• y₁ = - 2

• y₂ = - 5

Hence, Coordinates of E is evaluated as

$\rm :\longmapsto\:{\rm \: Coordinates \: of \: E = \bigg({ \dfrac{1 +2}{2} \; ,\; \dfrac{ - 2 - 5}{2} \bigg) \quad}}$

${\underline{\boxed{\rm \: Coordinates \: of \: E = \bigg({\dfrac{3}{2} \; ,\; \dfrac{ -7}{2} \bigg) \quad}}}}$

Now,

We know,

Equation of line passing through 2 points is given by

$\rm :\longmapsto\:\bf \:y - y_1 = \dfrac{y_2-y_1}{x_2-x_1} (x-x_1)$

Now,

$\rm :\longmapsto\:{\rm \: Coordinates \: of \: E = \bigg({\dfrac{3}{2} \; ,\; \dfrac{ -7}{2} \bigg) \quad}}$

$\rm :\longmapsto\:Coordinates \: of \: B = ( - 2, \: 3)$

Here,

$\rm :\longmapsto\:x_1 = - 2,y_1 = 3,x_2= \dfrac{3}{2} ,y_2= \dfrac{ - 7}{2}$

Hence, Equation of line BE is

$\rm :\longmapsto\:\:y - 3 = \dfrac{\dfrac{ - 7}{2} - 3}{\dfrac{3}{2} + 2 } (x + 2)$

$\rm :\longmapsto\:\:y - 3 = \dfrac{\dfrac{ - 7 - 6}{2}}{\dfrac{3 + 4}{2}} (x + 2)$

$\rm :\longmapsto\:\:y - 3 = \dfrac{ - 13}{7} (x + 2)$

$\rm :\longmapsto\:7y - 21 = - 13x - 26$

$\bf\implies \:13x + 7y + 5 = 0$

$\large{\boxed{\boxed{\bf{Option \: (b) \: is \: correct}}}}$

Additional Information :-

Different forms of equations of a STRAIGHT line

1. Equations of horizontal and vertical lines

Equation of the lines which are horizontal or parallel to the X-axis is y = a, where a is the y – coordinate of the points on the line.

Similarly, equation of a straight line which is vertical or parallel to Y-axis is x = a, where a is the x-coordinate of the points on the line

2. Point-slope form equation of line

Consider a non-vertical line L whose slope is m, A(x,y) be an arbitrary point on the line and P(a, b) be the fixed point on the same line. Equation of line is given by y - b = m(x - a)

3. Slope-intercept form equation of line

Consider a line whose slope is m which CUTS the Y-axis at a distance ‘a’ from the ORIGIN. Then the distance a is called the y– intercept of the line. The point at which the line cuts y-axis will be (0,a). Then equation of line is given by y = mx + a.

4. Intercept Form of Line

Consider a line L having x– intercept a and y– intercept b, then the line passes through X– axis at (a,0) and Y– axis at (0,b). Equation of line is given by x/a + y/b = 1.

5. Normal form of Line

Consider a perpendicular from the origin having length p to line L and it makes an angle β with the positive X-axis. Then, the equation of line is given by x cosβ + y sinβ = p.