Plot the following points on the Cartesian plane. a) A(-1,5) b) B(4,2) c) C(2,-2) d) D(-5,-3)

Mstari uliotolewa una mlinganyo $3y = 5 - 2x$.
Ili kupata gradient, tunahitaji kuupanga upya katika mfumo wa $y = mx + c$, ambapo $m$ ni gradient.

$$3y = -2x + 5$$

Gawanya pande zote mbili kwa 3:

$$y = \frac{-2}{3}x + \frac{5}{3}$$

Hivyo, gradient ya mstari huu, tuiite $m_1$, ni $-\frac{2}{3}$.

Mstari AB uko perpendicular (pembe mraba) kwa mstari uliotolewa.
Ikiwa mistari miwili iko perpendicular, bidhaa ya gradient zao ni -1.
Acha gradient ya mstari AB iwe $m_{AB}$.

$$m_{AB} \times m_1 = -1$$ $$m_{AB} \times \left(-\frac{2}{3}\right) = -1$$

Ili kutafuta $m_{AB}$, gawanya -1 kwa $-\frac{2}{3}$:

$$m_{AB} = \frac{-1}{-\frac{2}{3}}$$ $$m_{AB} = \frac{3}{2}$$

Mstari AB unapitia pointi A(2, 1) na B(4, m).
Gradient ya mstari AB inaweza pia kuhesabiwa kwa kutumia fomula $m = \frac{y_2 - y_1}{x_2 - x_1}$.
Kwa kutumia A(2, 1) kama $(x_1, y_1)$ na B(4, m) kama $(x_2, y_2)$:

$$m_{AB} = \frac{m - 1}{4 - 2}$$ $$m_{AB} = \frac{m - 1}{2}$$

Tunajua kutoka Hatua ya 2 kwamba $m_{AB} = \frac{3}{2}$.
Hivyo, tunaweza kulinganisha milinganyo miwili:

$$\frac{3}{2} = \frac{m - 1}{2}$$

Zidisha pande zote mbili kwa 2:

$$3 = m - 1$$

Ongeza 1 kwa pande zote mbili:

$$m = 3 + 1$$ $$\boxed{m = 4}$$

Tunayo gradient $m_{AB} = \frac{3}{2}$ na pointi A(2, 1).
Kwa kutumia mfumo wa pointi-gradient $y - y_1 = m(x - x_1)$:

$$y - 1 = \frac{3}{2}(x - 2)$$

Zidisha pande zote mbili kwa 2 ili kuondoa sehemu:

$$2(y - 1) = 3(x - 2)$$

Panua mabano:

$$2y - 2 = 3x - 6$$

Ongeza 2 kwa pande zote mbili:

$$2y = 3x - 6 + 2$$ $$2y = 3x - 4$$

Gawanya pande zote mbili kwa 2 ili kupata $y$ peke yake:

$$y = \frac{3}{2}x - \frac{4}{2}$$ $$\boxed{y = \frac{3}{2}x - 2}$$