The cartesian plane - class-XI

The Cartesian plane consists of two perpendicular and directed lines whose intersection point is the zero point for both the lines.

x - axis and y- axis intersect at the origin. SO, point of intersection is (0,0)

Since the x coordinate of the point is 0, its on the y axis.

The horizontal axis is called x-axis.

x and y coordinates are required to represent a point in cartesian plane.

i) The x-axis ,a line parallel to it is called horizontal line

The equation of line $y$=0 represents $x$ -axis ie.$(x,y)$ = $(x,0)$

$\dfrac {y _{2} - y _{1}}{x _{2} - x _{1}} = \dfrac {-4}{3} =\dfrac{3+5}{-5-x}=\dfrac{-4}{3}$

The smallest possible value of $\overline {AC} + \overline {BC}$ is obtained when $C$ is the intersection of the y-axis, with the line that leads from $A$ to the mirror image the mirror being the y-axis) $B' : (-2, 1)$ of $B$. This is true because $\overline {CB'} = \overline {CB}$ and a straight line is the shortest path between two points. The line through and $B'$ given by
$y = \dfrac {5 - 1}{5 + 2} x + k = \dfrac {4}{7} x + k$.
To find $k$, we use the fact that the line goes through $A$:
$5 = \dfrac {4}{7} . 5 + k, k = 5 - \dfrac {20}{7} = \dfrac {15}{7} = 2\dfrac {1}{7}$;
$\therefore C$ has coordinates $(0, 2\dfrac {1}{7})$.

The triangle cannot be equilateral if coordinates of vertices of the triangle is always rational.

Given 

For points A, B, and C to be collinear, the vectors AB and BC must be proportional. AB = (0-2, -1-1, 4-1) = (-2, -2, 3). BC = (K-0, 3-(-1), -2-4) = (K, 4, -6). Comparing components: -2/K = -2/4 = 3/-6. This gives -2/K = -1/2, so K = 4.

The line equation passing through (-4, 11) and (16, -1) is y - 11 = (( -1 - 11) / (16 - (-4))) * (x + 4), which simplifies to y - 11 = -12/20 * (x + 4) or 3x + 5y = 43. We look for positive integer solutions (x, y). Testing x values: if x=1, 5y=40, y=8; if x=6, 5y=25, y=5; if x=11, 5y=10, y=2. There are 3 such points.

The points are of the form (t, 2at + at^3). For these to be collinear, they must satisfy a linear equation. The sum of the abscissae (t1+t2+t3) for collinear points in this specific parametric form is 0.

This is a property of the intercept form of a line. For a line segment divided by a point P, the relationship between the segments and the intercepts leads to alpha + beta = 1.

Let the three points be,

Let the point (1, -1), $\displaystyle \left ( -\frac{1}{2},\frac{1}{2} \right )$ and (1, 2) be denoted by P, Q and R respectively
Now PQ = $\displaystyle \sqrt{\left ( -\frac{1}{2}-1 \right )^{2}+\left ( \frac{1}{2}+1 \right )^{2}}=\sqrt{\frac{18}{4}}=\frac{3}{2}\sqrt{2}$
QR = $\displaystyle \sqrt{\left ( 1+\frac{1}{2} \right )^{2}+\left ( 2-\frac{1}{2} \right )^{2}}=\sqrt{\frac{18}{4}}=\frac{3}{2}\sqrt{2}$
PR = $\displaystyle \sqrt{\left ( 1-1 \right )^{2}+\left ( 2+1 \right )^{2}}=\sqrt{9}=3 $
From the above we see that PQ = QR
$\displaystyle \therefore $ The triangle is isosceles

Put $x=5,,y=3$ in the given equation
LHS: $3x+2y=3\times5+2\times3$
$=15+6=21$
LHS $\neq RHS$
It does not lie on line $3x+2y=18$

1. A line parallel to x-axis is called a horizontal line parallel to y-axis is called a vertical line.
   2) The point of intersection between the X-axis and Y-axis is called the origin $(0,0)$

Given the point P has coordinate (3,4).Similarly O has (4,3)

Consider the given point

$A\left( 2a,4a \right),\,B\left( 2a,6a \right)\,and\,C\left( 2a+\sqrt{3}a,5a \right)$

Distance between AB,BC and CA  respectively.

$ AB=\sqrt{{{\left( 2a-2a \right)}^{2}}+{{\left( 4a-6a \right)}^{2}}} $

$ =\sqrt{0+{{(-2a)}^{2}}} $

$ =\sqrt{4{{a}^{2}}}=2a $

$ BC=\sqrt{{{\left( 2a-(2a+\sqrt{3}a) \right)}^{2}}+{{\left( 6a-5a \right)}^{2}}} $

$ =\sqrt{3{{a}^{2}}+{{a}^{2}}} $

$ =\sqrt{4{{a}^{2}}}=2a $

$ CA=\sqrt{{{\left( 2a+\sqrt{3}a-2a \right)}^{2}}+{{\left( 5a-4a \right)}^{2}}} $

$ =\sqrt{3{{a}^{2}}+{{a}^{2}}} $

$ =\sqrt{4{{a}^{2}}}=2a $

Hence, $AB=BC=CA$

It is an equilateral triangle.

Option (B) is correct answer.

The sequence of midpoints follows the rule A_n = (A_{n-1} + B) / 2. This is a geometric progression converging to B. After 10 steps, the coordinates are (1024 * (1 - (1/2)^10), 2050 * (1 - (1/2)^10)). Calculation: 1024 * (1023/1024) = 1023 and 2050 * (1023/1024) = 2046.009... which rounds to 2046.

$\begin{array}{l} Coordination, , of, , mid-po{ { int } }, , 0 \ =\left( { \frac { { 6+2 } }{ 2 } ,\frac { { 2-1 } }{ 2 }  } \right)  \ =\left( { 4,\frac { 1 }{ 2 }  } \right)  \ AB=BC \ \Rightarrow { \left( { { x _{ 1 } }-2 } \right) ^{ 2 } }+{ \left( { { y _{ 1 } }+1 } \right) ^{ 2 } }={ \left( { { x _{ 1 } }-6 } \right) ^{ 2 } }+{ \left( { { y _{ 1 } }-2 } \right) ^{ 2 } } \ \Rightarrow 8{ x _{ 1 } }+6{ y _{ 1 } }=35\to (i) \ AO=BO \ \Rightarrow { \left( { 2-4 } \right) ^{ 2 } }+{ \left( { -1-\frac { 1 }{ 2 }  } \right) ^{ 2 } }={ \left( { { x _{ 1 } }-4 } \right) ^{ 2 } }+{ \left( { { y _{ 1 } }-\frac { 1 }{ 2 }  } \right) ^{ 2 } } \ \Rightarrow 4+\frac { 9 }{ 4 } =x _{ _{ 1 } }^{ 2 }+y _{ 1 }^{ 2 }-8{ x _{ 1 } }-{ y _{ 1 } }+16+\frac { 1 }{ 4 }  \ \Rightarrow x _{ 1 }^{ 2 }+y _{ 1 }^{ 2 }-8{ x _{ 1 } }-{ y _{ 1 } }=-10\to (ii) \ from, , equation, , (i) \ put \ { x _{ 1 } }=\frac { { 35-6{ y _{ 1 } } } }{ 8 } , , in, , equation, , (ii) \ \Rightarrow { \left( { \frac { { 35-6{ y _{ 1 } } } }{ 8 }  } \right) ^{ 2 } }+y _{ 1 }^{ 2 }-\left( { 35-6{ y _{ 1 } } } \right) { y _{ 1 } }=-10 \ \Rightarrow 4y _{ 1 }^{ 2 }-4{ y _{ 1 } }-15=0 \ \Rightarrow \left( { 2{ y _{ 1 } }+3 } \right) \left( { { y _{ 1 } }-5 } \right) =0 \ \Rightarrow { y _{ 1 } }=\frac { { -3 } }{ 2 } ,5 \ { y _{ 1 } }=\frac { { -3 } }{ 2 } \to { x _{ 1 } }=\frac { { 35-6\times -\frac { 3 }{ 2 }  } }{ 8 } =\frac { { 11 } }{ 2 }  \ { y _{ 1 } }=5\to { x _{ 1 } }=\frac { { 35-6\times 5 } }{ 8 } =\frac { 5 }{ 8 }  \ The, , vertices, of, , other, , two, vertices, , are \ \left( { \frac { { 11 } }{ 2 } ,\frac { { -3 } }{ 2 }  } \right) , , and, , \left( { \frac { 5 }{ 8 } ,5 } \right)  \end{array}$