(i) write down the co-ordinates of the point p that divides the line joining a(-4, 1) and b(17, 10) in the ratio 1 : 2. (ii) calculate the distance op, where o is the origin. - section formula
(i) write down the co-ordinates of the point p that divides the line joining a(-4, 1) and b(17, 10) in the ratio 1 : 2.(ii) calculate the distance op, where o is the origin.(iii) in what ratio does the y-axis divide the line ab?
(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.
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 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. ?solution 
step 1:
we have to draw the graph of form r to r. 
step 2:
the graph of f is the set of all points (x,f(x)) with x in the domain of f.
we take x in [-4,4]
and we will find the separate values of y for separate intervals.
 
 
 

 
......(1)
 

 

 

using x values along x-axis and along y-axis we get the graph of given function form r to r.
step 3:
the graph is as follow slope of the line ⇒ slope of the lineequation of line :⇒ 3(y - 5) = 2(x - 2)⇒ 2x - 3y + 11 = 0 …equation (i)putting x = x + 3 and y = y + 2 in the above equation we get2(x + 3) - 3(y + 2) + 11 = 0⇒ 2x - 3y + 11 = 0since it again gives the same equation as (i) so it is a point on this line.hence proved the straight distance between two place on a map can be measured by a strip of paper and divider.use a ruler to measure the distance between the two places. if the line is quite curved, use a string to determine the distance, and then measure the string. find the scale for the map you're going to use. it might be a ruler bar scale or a written scale, in words or numbers. you can find the radius in the following way. 
write the equation of the line that is perpendicular to the given tangent line. this perpendicular goes through the stated center, and intersects the tangent; the perpendicular's slope is the negative reciprocal of the stated tangent line.
find the intersection (point t) of the tangent and radius line--then use the distance formula to find the radius, which is a segment from the stated center to the point t. given: (i) xy = 4.5cm

(ii) yz = 5 cm

(iii) zx = 6 cm

to construct: a xyz

steps for construction

step 1 : first we will draw a line segment xy of length 4.5cm



step 2 : we have to draw an arc , taking x as center and radius 6cm



step 3 : we have to another arc, with radius 5cm and center at y and mark the intersection of arcs as z.



step 4 : we have to join xz and xy



this is the required triangle 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) given the boundaries of the area to be found are,• the curve • the x-axis• x = 0 (y-axis)• x = 4 (a line parallel toy-axis)as per the given boundaries,• the curve,is a curve with vertex at .• x=2 is parallel toy-axis at 2 units away from the y-axis.• x=0 is the y-axis.• the four boundaries of the region to be found are,•point a, where the curve y2 = 6x + 4 and x=0 meet.•point b, where the curve y2 = 6x + 4 and x=2 meet.•point c, where the x-axis and x=2 meet i.e. c(2,0).•point o, or the origin i.e. o(0,0).area of the required region = area of oabc.[using the formula ]the area of the required region you are correct.there is a simple mathematical reason why your answer is correct, but wyzant told me that adding it to this answer required review by wyzant.try again:p(a or b) = p(a) + p(b) - p(a and b)p(neither a nor b) = 1 - p(a or b)