Math, asked by BrainlyPhantom, 1 day ago

Coordinate Geometry
1. Find the ratio in which the line 3x+4y=28 divides the line segment joining the points (1,3) and (2,7).
2. Let the opposite angular points of a square be (3,4) and (1,−1). Find the coordinates of the remaining angular points.
Well-explained answers required. Thank you in advance. :)

Answers

Answered by anindyaadhikari13

$\textsf{\large{\underline{Solution 1}:}}$

Let the coordinate of the point in which the line 3x + 4y = 28 the line divides the line segment be P(x, y).

Let the ratio in which the line divides AB be k : 1.

Given Points: (1, 3) and (2, 7)

Here:

$:\longmapsto \begin{cases}\rm x_{1}=1\\ \rm x_{2}=2\\ \rm y_{1}=3\\ \rm y_{2}=7\end{cases}$

Then, by using section formula, the coordinates of P will be:

$\rm=(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}},\dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}})$

$\rm=\bigg(\dfrac{2k+1}{k+1},\dfrac{7k+3}{k+1}\bigg)$

This is the point through which the line 3x + 4y = 28 passes. Therefore, it must be a solution of the equation 3x + 4y = 28.

Substitute the value in the formula, we get:

$\rm:\longmapsto3\times\dfrac{2k+1}{k+1}+4\times\dfrac{7k+3}{k+1}=28$

$\rm:\longmapsto \dfrac{6k+3}{k+1}+\dfrac{28k+12}{k+1}=28$

$\rm:\longmapsto \dfrac{34k+15}{k+1}=28$

$\rm:\longmapsto34k+15=28(k+1)$

$\rm:\longmapsto34k+16=28k+28$

$\rm:\longmapsto34k-28k=28-15$

$\rm:\longmapsto 6k=13$

$\rm:\longmapsto k=\dfrac{13}{6}$

Therefore, the ratio in which it divides is k : 1 i.e.:

= 13/6 : 1

= 13 : 6

$\textsf{\large{\underline{Solution 2}:}}$

Let ABCD be a square where coordinates A = (3, 4) and that of C = (1, -1)

Let the coordinates of B be (x, y).

As it is a square:

$\rm:\longmapsto AB=BC$

By using distance formula:

$\rm:\longmapsto\sqrt{(x-3)^{2}+(y-4)^{2}}=\sqrt{(x-1)^{2}+(y+1)^{2}}$

Squaring both sides, we get:

$\rm:\longmapsto(x-3)^{2}+(y-4)^{2}=(x-1)^{2}+(y+1)^{2}$

On Simplifying the given equation, we get:

$\rm:\longmapsto x^{2}-6x+9+y^{2}-8y+16=x^{2}-2x+1+y^{2}+2y+1$

$\rm:\longmapsto-6x+9-8x+16=-2x+1+2y+1$

$\rm:\longmapsto 4x+10y-23=0$

$\rm:\longmapsto x=\dfrac{23-10y}{4} -(i)$

Now, in ΔABC, we apply Pythagoras Theorem:

$\rm:\longmapsto AB^{2}+BC^{2}=(AC)^{2}$

$\rm:\longmapsto(x-3)^{2}+(y-4)^{2}+(x-1)^{2}+(y+1)^{2}=(3-1)^{2}+(4+1)^{2}$

On simplifying the given equation, we get:

$\rm:\longmapsto x^{2}-6x+9+y^{2}-8y+16+x^{2}-2x+1+y^{2}+2y+1=29$

$\rm:\longmapsto 2x^{2}-8x+2y^{2}-6y+27=29$

$\rm:\longmapsto 2x^{2}-8x+2y^{2}-6y=2$

$\rm:\longmapsto x^{2}+y^{2}-4x-3y-1=0$

From (i):

$\rm:\longmapsto \bigg(\dfrac{23-10y}{4}\bigg)^{2}+y^{2}-(23-10y)-3y-1=0$

On simplifying, we get:

$\rm:\longmapsto4y^{2}-12y+5=0$

Factorising it, we get:

$\rm:\longmapsto (2y-1)(2y-5)=0$

Therefore:

$\rm:\longmapsto y=\dfrac{1}{2},\dfrac{5}{2}$

So from (i), the values of x will be:

$\rm:\longmapsto x=\dfrac{9}{2},\dfrac{-1}{2}$

⊕ Therefore, the other coordinates are - (9/2, 1/2) and (5/2, -1/2).

$\textsf{\large{\underline{Learn More}:}}$

1. Section formula.

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the point which divides PQ internally in the ratio m₁ : m₂. Then, the coordinates of R will be:

$\rm:\longmapsto R = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)$

2. Mid-point formula.

Let P(x₁, y₁) and Q(x₂, y₂) be two points in the coordinate plane and R(x, y) be the mid-point of PQ. Then, the coordinates of R will be:

$\rm:\longmapsto R = \bigg(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\bigg)$

3. Centroid of a triangle.

Centroid of a triangle is the point where the medians of the triangle meet.

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle. Let R(x, y) be the centroid of the triangle. Then, the coordinates of R will be:

$\rm:\longmapsto R = \bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3}, \dfrac{y_{1}+y_{2}+y_{3}}{3}\bigg)$