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...

The directions of the centroid are just the normal of the directions of the vertices. So to track down the x facilitate of the orthocenter, include the three vertex x organizes and gap by three....

The directions of the centroid are just the normal of the directions of the vertices. So to track down the x facilitate of the orthocenter, include the three vertex x organizes and gap by three....

The centroid is the middle place of the item. The point in which the three medians of the triangle converge is known as the centroid of a triangle. It is likewise characterized as the mark of...

The directions of the centroid are just the normal of the directions of the vertices. So to track down the x facilitate of the orthocenter, include the three vertex x organizes and gap by three....

As we know that the coordinates of the centroid of a triangle whose vertices are $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right)$Are $\left(...

A diagram comprises of two tomahawks called the x (even) and y (vertical) tomahawks. These tomahawks compare to the factors we are relating. In financial aspects we will generally give the tomahawks...

A diagram comprises of two tomahawks called the x (even) and y (vertical) tomahawks. These tomahawks compare to the factors we are relating. In financial aspects we will generally give the tomahawks...