As we know that (x,y) is the mid-point $x=(3+k)/2$ and $y=(4+7)/2=11/2$ Also it is given that the mid-point lies on the line $2x+2y+1=0$ $2[(3+k)/2]+2(11/2)+1=0$ $3+k+11+1=0$ Thus, $k=-15$

Let’s consider P(x, y) be the required point. By using section formula, we know that the coordinates are $x=\frac{m{{x}_{2}}+n{{x}_{1}}}{m+n}$ $y=\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n}$ Here,...

Let’s consider A$(3,-2)$, B$(4,0)$,C$(6,-3)$ and D$(5,-5)$ Let’s take P(x, y) be the point of intersection of diagonals AC and BD of ABCD. The mid-point of AC is provided that,...

Let’s consider A$(-2,-1)$,B$(1,0)$,C$(4,3)$ and D$(1,2)$ are the given points. Let’s take P(x, y) be the point of intersection of the diagonals of the parallelogram formed by the given points. As We...

Given that, P($9a–2$, -b) divides the line segment joining A$(3a+1,-3)$ and B$(8a,5)$ in the ratio$3:1$ Therefore, by using section formula The Coordinates of P are $9a-2=\frac{3(8a)+1(3a+1)}{3+1}$...

As it is given (a, b) is the mid-point of the line segment A($10,-6$) and B(k,$4$) Therefore, (a, b) $=(10+k/2,-6+4/2)$ a $=(10+k)/2$ and b $=-1$ $2a=10+k$ $K=2a–10$ Given that, $a–2b=18$ By Using...

Let’s consider the point P($2$, y) divide the line segment joining the points A$(-2,2)$ and B$(3,7)$ in the ratio k: 1 Now, the coordinates of P are given by $\left[ \frac{3k+(-2)\times...

Let’s consider AD be the median through A. As we know that, AD is the median and D is the mid-point of BC Therefore, the coordinates of D are $(1+5/2,-1+1/2)=(3,0)$ So, Length of median...

Let’s consider P$(-2,-3)$ and Q$(9,3)$ be the given points. Let’s Suppose we have the y-axis that divides PQ in the ratio k:$1$ at R($0$, y) So, the coordinates of R are as given below Now, on...

Given that the points are A$(-3,2)$, B$(-5,5)$, C$(2,-3)$ and D$(4,4)$ So, Coordinates of the mid-point of AC are $(-3+2/2,2-3/2)=(-1/2,-1/2)$ And, The Coordinates of mid-point of BD are...

Given that, Points A$(1/2,3/2)$ and B$(2,-5)$ Let’s consider the point P$(3/4,5/12)$ divide the line segment AB in the ratio k:$1$ As, we know that P$(3/4,5/12)=(2k+1/2)/(k+1),(2k+3/2)/(k+1)$...

Let’s A $(-2,-3)$ and B$(5,6)$ be the given points. (i) Suppose that x-axis divides AB in the ratio k:$1$ at the point P Now, the coordinates of the point of division are $\left[...

Let’s A$(4,5)$, B$(7,6)$,C$(6,3)$ and D$(3,2)$ be the given points. And, P be the point of intersection of AC and BD. Coordinates of the mid-point of AC are $(4+6/2,5+3/2)=(5,4)$ Coordinates of the...

Let’s A$(4,3)$,B$(6,4)$,C$(5,6)$ and D$(3,5)$ be the given points. We know the distance formula is $D=\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}}$...

Let’s A$(-4,-1)$,B$(-2,-4)$,C$(4,0)$ and$(2,3)$ be the given points. Now we have, Coordinates of the mid-point of AC are $(-4+4/2,-1+0/2)$ =$(0,-1/2)$ Coordinates of the mid-point of BD are...

Let’s AD, BF and CE be the medians of ΔABC The Coordinates of D are $(5+1/2,1–1/2)$ $=(3,0)$ Coordinates of E are $(-1+1/2,3–1/2)$ $=(0,1)$ Coordinates of F are $(5–1/2,1+3/2)$ $=(2,2)$ Now, Finding...

Let’s the point on the x-axis be (x, $0$). [y – coordinate is zero] And, let’s this point divides the line segment AB in the ratio of k :$1$. Now by using the section formula for the y-coordinate,...

Let’s P divide the line joining A and B and let it divide the segment in the ratio k:$1$ Now, by using the section formula for the y – coordinate we have $2=(-3k+5)/(k+1)$ $2(k+1)=-3k+5$...

Let’s P divide A$(-3,10)$ and B$(6,-8)$ in the ratio of k:$1$ Given that the coordinates of P as ($-1$,y) Now, by using the section formula for x – coordinate we have $-1=6k–3/k+1$ $-(k+1)=6k–3$...

Given that the points are A$(2,0)$, B$(9,1)$, C$(11,6)$ and D$(4,4)$. Now Coordinates of mid-point of AC are $(11+2/2,6+0/2)=(13/2,3)$ Coordinates of mid-point of BD are $(9+4/2,1+4/2)=(13/2,5/2)$...

Let’s the point $(-4,6)$ divide the line segment AB in the ratio k:$1$. Thus, by using the section formula, we have $(-4,6)=\left( \frac{3k-6}{k+1},\frac{-8k+10}{k+1} \right)$ $-4=\frac{3k-6}{k+1}$...

Let’s A $(-2,1)$, B$(1,0)$, C$(x,3)$ and D$(1,y)$ be the given points of the parallelogram. As We know that the diagonals of a parallelogram bisect each other. Therefore, the coordinates of...

Let’s the coordinates of point A be (x, y) If we have AB is the diameter, then the center in the mid-point of the diameter Thus , $(2,-3)=(x+1/2,y+4/2)$ $2=x+1/2$ and $-3=y+4/2$ $4=x+1$ and $-6=y+4$...

Let’s P$(5,-6)$ and Q$(-1,-4)$ be the given points. Let’s the y-axis divide the line segment PQ in the ratio k: $1$ Now, by using section formula for the x-coordinate (as it’s zero) Now we have...