if p(2,p) is the mid-point of the line segment joining the points a(6, -5)and b(-2, 11), find the value of p
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you seem to be missing some information about e.if e is the midpoint between d and f, thene is (1,1) take half of (4-2) = 1 and half of (5-3) = 1e alphabetically does come between d and f.just average the x values and average the y values of the two points d and for plot them graphically, carefully, maybe on some graph paper, and you'llvisually see the midpoint is at (1,1) (i) co-ordinates of point p are `((1xx17+2xx(-1))/(1+2),(1xx10+2xx1)/(1+2))` `=((17-8)/3,(10+2)/3)` `=(9/3,12/3)` `=(3,4)` (ii) `op=sqrt((0-3)^2+(0-4)^2)` `op=sqrt(9+16)` `op=sqrt25` `op=5` units (iii) let ab be divided by the point p(0, y) lying on y-axis in the ratio `therefore (0,y)=((kxx17+1xx(-4))/(k+1),(kxx10+1xx1)/(k+1))` `=>(0,y)=((17k-4)/(k+1),(10k+1)/(k+1))` `=>0=(17k-4)/(k+1)` `=>17k-4=0` `=>k=4/17` thus, the ratio in which the y-axis divide the line ab is 4: 17. several ways to do this, here's one:

to go from (-6,4) to (-2,5) you need to go 4 in the x direction and 1 in the y direction. if you move again 4 in the x direction and 1 in the y direction you end up at (2, 6). that's p2 in this we are asked to divide it in 2:3 ratioso, 3+2 = 55, number congruent of line segment is formed.i. firstly, we have to draw a line segment ab of length 7 cm. draw a line with an acute angle from point aii. point the compass on a, and cut the new line with some radiusplace the compass along the line and make 5 arcs. since we need 5 congruent parts, and name the last one as “g”iii. take length of bg as radius draw an arc from a just below it.again, take the length of ag as radius and draw an arc which cuts the previous arc from b, name the intersection point as hconnect b and hiv. taking the same length that we used in line ag, draw arcs along bh and name the points.v. connect the points along ag and bh. name the intercepts on the line ab as m, n, o & p.vi. thus we can divide a line segment as ao:ob = 6:4= 3:2 solution: ` x = {:((-4/5,-1),(2/5,1)):}` find zeroes of the following polynomials (with steps ) (i) p(x)=6x^2-3-7x (ii) f(x)=2-x^2 hint: assume the co-ordinates of $r$ as $(x,y)$ and as it is given that $p,q,r$ are collinear points. apply distance condition of collinear points which is if $p,q,r$ are three collinear points then they must satisfy $pq+qr=pr$ and get one equation. apply distance formula which states that if $a({{x}\_{1}},{{y}\_{1}})$ and $b({{x}\_{2}},{{y}\_{2}})$ are two points then the distance $ab=\sqrt{{{({{x}\_{2}}-{{x}\_{1}})}^{2}}+{{({{y}\_{2}}-{{y}\_{1}})}^{2}}}$ and get the second equation by using $pr=10$ , solve these two equations and get the co-ordinates of $r$. complete step-by-step solution:it is given in the question that $p, q, r$ are collinear points.collinear means lying on the same line which means that $p, q, r$ lies on the same straight line.let us suppose that the co-ordinates of $r$ is $(x,y)$. now, we have $p(7,7)$, $q(3,4)$, $r(x,y)$ and also $pr=10$.we have to use distance formula in order to find $pq,qr,pr$. so, according to distance formula if $a({{x}\_{1}},{{y}\_{1}})$ and $b({{x}\_{2}},{{y}\_{2}})$ then $ab=\sqrt{{{({{x}\_{2}}-{{x}\_{1}})}^{2}}+{{({{y}\_{2}}-{{y}\_{1}})}^{2}}}$.now, $p,q,r$ are three collinear points then they must satisfy,$\therefore pq+qr=pr$using distance formula, we get$\rightarrow \sqrt{{{(7-3)}^{2}}+{{(7-4)}^{2}}}+\sqrt{{{(3-x)}^{2}}+{{(4-y)}^{2}}}=10$$\rightarrow \sqrt{16+9}+\sqrt{{{(3-x)}^{2}}+{{(4-y)}^{2}}}=10$$\rightarrow 5+\sqrt{{{(3-x)}^{2}}+{{(4-y)}^{2}}}=10$\[\rightarrow \sqrt{{{(3-x)}^{2}}+{{(4-y)}^{2}}}=5\]squaring both sides, we get$\rightarrow {{(3-x)}^{2}}+{{(4-y)}^{2}}=25$$\rightarrow {{x}^{2}}+{{y}^{2}}-6x-8y+25=25$cancelling 25 from both sides, we get$\rightarrow {{x}^{2}}+{{y}^{2}}=6x+8y...................(1)$also, it is given that $pr=10$, using this we get$\therefore pr=10$\[\rightarrow \sqrt{{{(7-x)}^{2}}+{{(7-y)}^{2}}}=10\]squaring both sides, we get\[\rightarrow {{(7-x)}^{2}}+{{(7-y)}^{2}}=100\]$\rightarrow {{x}^{2}}+{{y}^{2}}-14x-14y+98=100$putting the value of ${{x}^{2}}+{{y}^{2}}$ from equation (1), we get$\rightarrow 6x+8y-14x-14y=100-98$$\rightarrow -8x-6y=2$now, expressing $y$ in terms of $x$, we get$\rightarrow y=\dfrac{-(2+8x)}{6}=\dfrac{-(1+4x)}{3}..................(2)$putting the value of $y$ from equation (2) in equation (1), we get$\rightarrow {{x}^{2}}+{{\left( \dfrac{-(1+4x)}{3} \right)}^{2}}=6x+\dfrac{-(1+4x)}{3}\times 8$$\rightarrow \dfrac{9{{x}^{2}}+1+8x+16{{x}^{2}}}{9}=\dfrac{18x-8-32x}{3}$$\rightarrow 25{{x}^{2}}+8x+1=3(-14x-8)$$\rightarrow 25{{x}^{2}}+8x+1+42x+24=0$$\rightarrow 25{{x}^{2}}+50x+25=0$$\rightarrow 25({{x}^{2}}+2x+1)=0$$\rightarrow {{(x+1)}^{2}}=0$$\rightarrow x+1=0\rightarrow x=-1$now, putting the value of $x$ in equation (2), we get$\rightarrow y=\dfrac{-(1+4\times -1)}{3}$$\rightarrow y=\dfrac{3}{3}=1$so, $x=-1$ and $y=1$.hence the co-ordinates of $r$ is $(-1,1)$. note: this question is simple and there are many other alternative ways to solve it but the tricky part is to choose which method to solve the question. there are three different conditions of collinearity as there is slope condition, distance condition, and area condition. as in the other two conditions distances are not involved, only the distance condition involves the use of distances. in the question, distance is mentioned so we choose that method to solve it. hence, students should carefully read the question and choose which method is suitable to solve the question according to the values given. 1/
a/ the x-intersect between 2 graphs , x^2+4x-5=0>x = -5 or 1
v = pi { integral of [(4x)^2 - (5-x^2)^2] dx , from x=-5 to x=1 } if , passing through (1, 2), satisfies the differential equation , then multiply 2 in both lhs and rhs,integrating on both the sides,