Step-by-step explanation:
3
Let D be the mid point of side BC, [B(2, - 1), C(5, 2)].
Therefore, by mid-point formula:
[tex]D = ( \frac{2 + 5}{2}, \: \: \frac{ - 1 + 2}{2} ) = ( \frac{7}{2}, \: \: \frac{ 1}{2} ) \\ \therefore D= (3.5, \: \: 0.5) \\ \& \: A=(-1,\:\:4)...(given) \\\\ now \: by \: distance \: formula \\ \\ Length \: of \: segment \: AD \\ = \sqrt{( - 1 - 3.5)^{2} + {(4 - 0.5)}^{2} } \\ = \sqrt{(4.5)^{2} + {(3.5)}^{2} } \\ = \sqrt{20.25 + 12.25 } \\ = \sqrt{32.5} \\ \red{ \boxed{\therefore Length \: of \: segment \: AD = 5.7 \: units}}[/tex]
4 (a)
Equation of line AB[A(2, 1), B(-2, - 11)] in two point form is given as:
[tex]\frac{y-y_1}{y_1-y_2} =\frac{x-x_1}{x_1 - x_2} \\\\
\therefore \frac{y-1}{1-(-11)} =\frac{x-2}{2 - (-2) } \\\\
\therefore \frac{y-1}{1+11} =\frac{x-2}{2 +2} \\\\
\therefore \frac{y-1}{12} =\frac{x-2}{4} \\\\
\therefore \frac{y-1}{3} =\frac{x-2}{1} \\\\
\therefore y-1= 3(x - 2)\\\\
\therefore y= 3x - 6+1\\\\
\therefore y= 3x - 5\\\\
\huge \purple {\boxed {\therefore 3x - y-5=0}} \\ [/tex]
is the equation of line AB.
Now we have to check whether C(4, 7) lie on line AB or not.
Let us substitute x = 4 & y = 7 on the Left hand side of equation of line AB and if it gives us 0, then C lies on the line.
[tex] LHS = 3x - y-5\\
=3\times 4-7-5\\
= 12-12\\
=0\\
= RHS[/tex]
Hence, point C (4, 7) lie on the straight line AB.
4(b)
Like we did in 4(a), first find the equation of line AB and then substitute the coordinates of point C in equation and if they satisfy the equation, then all the three points lie on the straight line.
